Integral Calculus jee mains questions | JEE Main Maths Integral Calculus Previous Year Question

Q 31 :

The area of the region bounded by parabola $y^{2} = x$ and the straight line $2 y = x$ is:

$\frac{4}{3}$ sq. units
$1$ sq. unit
$\frac{2}{3}$ sq. unit
$\frac{1}{3}$ sq. unit

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Ans. $\frac{4}{3}$ sq. units

Explanation :

We have to find the area enclosed by parabola $y^{2} = x$ and the straight line $2 y = x .$

$∴$ ${(\frac{x}{2})}^{2} = x$

$\Rightarrow$ $x^{2} = 4 x$

$\Rightarrow$ $x (x - 4) = 0$

$\Rightarrow$ $x = 4$

$\Rightarrow$ $y = 2$

and $x = 0$

$\Rightarrow$ $y = 0$

So, the intersection points are (0, 0) and (4, 2).

Area enclosed by shaded region,

$A = \int_{0}^{4} (\sqrt{x} - \frac{x}{2}) d x$

$= {[\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - \frac{1}{2} \cdot \frac{x^{2}}{2}]}_{0}^{4}$

$= {[2 \cdot \frac{x^{3 / 2}}{3} - \frac{x^{2}}{4}]}_{0}^{4}$

$= \frac{2}{3} {(4)}^{3 / 2} - \frac{16}{4} - \frac{2}{3} (0) + \frac{1}{4} (0)$

$= \frac{16}{3} - \frac{16}{4} = \frac{64 - 48}{12} = \frac{16}{12} = \frac{4}{3} sq. units.$

Q 32 :

Area bounded by the parabola $y = 4 x^{2},$ Y-axis and the lines $y = 1, y = 4$ is:

3 sq. units
$\frac{7}{5}$ sq. units
$\frac{7}{3}$ sq. units
None of these

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Ans. $\frac{7}{3}$ sq. units

Explanation :

Given, $y = 4 x^{2}$

Area to be found between Y-axis, $y = 1$ and $y = 4$ .

The figure is as follows:

At $y = 1$ $\to y = 4 x^{2}$ gives $x = \frac{1}{2}$ and $- \frac{1}{2}$

At $y = 4$ $\to y = 4 x^{2}$ gives $4 = 4 x^{2}$ or $x = 1, - 1$

Here,

$Area = \int_{1}^{4} x d y = \int_{1}^{4} \frac{\sqrt{y}}{2} d y$

$= \frac{1}{2} \int_{1}^{4} y^{1 / 2} d y$

$= \frac{1}{2} {[\frac{2}{3} y^{3 / 2}]}_{1}^{4}$ $= \frac{1}{3} {[y^{3 / 2}]}_{1}^{4}$

$= \frac{1}{3} [4^{3 / 2} - 1^{3 / 2}]$

$= \frac{1}{3} (8 - 1) = \frac{7}{3} sq. units.$

Q 33 :

The area bounded by the parabola $y^{2} = 4 a x,$ its axis and two ordinates $x = 4, x = 9$ is:

$4 a^{2}$ sq. units
$4 a^{2} \cdot 4$ sq. units
$4 a^{2} (9 - 4)$ sq. units
$\frac{152 \sqrt{a}}{3}$ sq. units

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Ans. $\frac{152 \sqrt{a}}{3}$ sq. units

Explanation :

$Shaded area A = 2 \int_{4}^{9} \sqrt{4 a x} d x$

$A = 4 \sqrt{a} \int_{4}^{9} \sqrt{x} d x$

$= 4 \sqrt{a} \times \frac{2}{3} {[x^{3 / 2}]}_{4}^{9}$

$= \frac{8 \sqrt{a}}{3} (9^{3 / 2} - 4^{3 / 2})$

$= \frac{8 \sqrt{a}}{3} (27 - 8)$

$= \frac{152 \sqrt{a}}{3} sq. units.$

Q 34 :

Area bounded by the parabola $y^{2} = 4 a x$ and its latus-rectum is:

$\frac{2}{3} a^{2}$ sq. unit
$\frac{4}{3} a^{2}$ sq. unit
$\frac{8}{3} a^{2}$ sq. unit
$\frac{3}{8} a^{2}$ sq. unit

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Ans. $\frac{8}{3} a^{2}$ sq. unit

Explanation:

$Area = 2 \int_{0}^{a} y d x = 2 \int_{0}^{a} \sqrt{4 a x} d x$

$= 2 \times 2 \sqrt{a} \int_{0}^{a} \sqrt{x} d x$

$= 4 \sqrt{a} \times \frac{2}{3}$ $| x^{3 / 2} |_{0}^{a}$

$= \frac{8 \sqrt{a}}{3} (a^{3 / 2} - 0)$

$= \frac{8}{3} a^{2} sq. units.$

Q 35 :

The area between the curve $y^{2} = 4 a x$ , the $x$ -axis and the lines $x = 0$ and $x = a$ is:

$\frac{4}{3} a^{2}$ sq. units
$\frac{8}{3} a^{2}$ sq. units
$\frac{2}{3} a^{2}$ sq. units
$\frac{5}{3} a^{2}$ sq. units

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Ans. $\frac{8}{3} a^{2}$ sq. units

Explanation:

$Required area = 2 \int_{0}^{a} y d x$

$= 2 \int_{0}^{a} \sqrt{4 a x} d x$

$= 4 \sqrt{a} \int_{0}^{a} x^{1 / 2} d x$

$= 4 \sqrt{a} \times \frac{2}{3} {[x^{3 / 2}]}_{0}^{a}$

$= \frac{8 \sqrt{a}}{3} \cdot a \sqrt{a}$

$= \frac{8}{3} a^{2} sq. units.$

Q 36 :

The area bounded by the curve $y = x^{2}$ , the $Y$ -axis and the $X$ -axis and $x = 3$ is:

5 sq. units
7 sq. units
9 sq. units
10 sq. units

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Ans. 9 sq. units

Explanation:

$The required area, A = \int_{x = 0}^{3} y \cdot d x$

$A = \int_{0}^{3} x^{2} \cdot d x = {[\frac{x^{3}}{3}]}_{0}^{3}$

$= 9 - 0$

$= 9 sq. units.$

Q 37 :

The area of the regions bounded by the curve $y = x^{2}$ , the $X$ -axis and the lines $x = 1$ and $x = 3$ is:

$\frac{26}{3}$ sq. units
$\frac{14}{5}$ sq. units
$\frac{26}{5}$ sq. units
$\frac{14}{3}$ sq. units

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Ans. $\frac{26}{3}$ sq. units

Explanation:

$Let A be the required area.$

$A = \int_{1}^{3} y d x$

$= \int_{1}^{3} x^{2} d x$

$= \frac{1}{3} {[x^{3}]}_{1}^{3}$

$= \frac{1}{3} (27 - 1)$

$A = \frac{26}{3} sq. units.$

Q 38 :

The area of the regions bounded by the curve $y^{2} = 4 x$ , the $X$ -axis and the lines $x = 1$ , $x = 4, y \geq 0$ is:

$\frac{26}{3}$ sq. units
$\frac{14}{5}$ sq. units
$\frac{28}{3}$ sq. units
$\frac{14}{3}$ sq. units

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Ans. $\frac{28}{3}$ sq. units

Explanation:

$Let A be the required area.$

$A = \int_{1}^{4} y d x$

$= \int_{1}^{4} 2 \sqrt{x} d x$

$= 2 \int_{1}^{4} x^{1 / 2} d x$

$= 2 \cdot \frac{2}{3} {[x^{3 / 2}]}_{1}^{4}$

$= \frac{4}{3} [4^{3 / 2} - 1]$

$∴$ $A = \frac{28}{3} sq. units.$

Q 39 :

The area of the given bounded by the curves $x^{2} = 16 y$ , $y = 1$ , $y = 4$ and the $Y$ -axis, lying in the first quadrant:

$\frac{26}{3}$ sq. units
$\frac{28}{3}$ sq. units
$\frac{56}{3}$ sq. units
$\frac{14}{3}$ sq. units

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Ans. $\frac{56}{3}$ sq. units

Explanation:

$Let A be the required area.$

$A = \int_{1}^{4} x d y$

$= \int_{1}^{4} \sqrt{16 y} d y = \int_{1}^{4} 4 \sqrt{y} d y$

$= 4 {[\frac{2}{3} y^{3 / 2}]}_{1}^{4}$

$= \frac{8}{3} \times [8 - 1] = \frac{8}{3} \times 7$

$∴$ $A = \frac{56}{3} sq. units.$

Q 40 :

The area of the region bounded by the curve $x^{2} = 4 y$ and the straight line $x = 4 y - 2$ is:

$\frac{3}{8}$ sq. unit
$\frac{5}{8}$ sq. unit
$\frac{7}{8}$ sq. unit
$\frac{9}{8}$ sq. units

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Ans. $\frac{9}{8}$ sq. units

Explanation:

Given equation of curve is $x^{2} = 4 y$ and the straight line is $x = 4 y - 2 .$

For the point of intersection, put $x = 4 y - 2$ in the equation of the curve:

${(4 y - 2)}^{2} = 4 y$

$\Rightarrow$ $16 y^{2} + 4 - 16 y = 4 y$

$\Rightarrow$ $16 y^{2} - 20 y + 4 = 0$

$\Rightarrow$ $4 y^{2} - 5 y + 1 = 0$

$\Rightarrow$ $4 y^{2} - 4 y - y + 1 = 0$

$\Rightarrow$ $4 y (y - 1) - 1 (y - 1) = 0$

$\Rightarrow$ $(4 y - 1) (y - 1) = 0$

$∴$ $y = 1, \frac{1}{4}$

For $y = 1$ , $x = \sqrt{4 \cdot 1} = 2$

For $y = \frac{1}{4}$ , $x = \sqrt{4 \cdot \frac{1}{4}} = - 1$

So, the intersection points are $(2, 1)$ and $(- 1, \frac{1}{4})$

Hence, the area of the shaded region is

$= \int_{- 1}^{2} (\frac{x + 2}{4}) d x - \int_{- 1}^{2} \frac{x^{2}}{4} d x$

$= \frac{1}{4} {[\frac{x^{2}}{2} + 2 x]}_{- 1}^{2} - \frac{1}{4} {[\frac{x^{3}}{3}]}_{- 1}^{2}$

$= \frac{1}{4} [\frac{4}{2} + 4 - \frac{1}{2} + 2] - \frac{1}{4} [\frac{8}{3} + \frac{1}{3}]$

$= \frac{1}{4} \cdot \frac{15}{2} - \frac{1}{4} \cdot \frac{9}{3} = \frac{45 - 18}{24}$

$= \frac{27}{24} = \frac{9}{8} sq. units.$

Q 41 :

If the area bounded by the curve $y^{2} = 4 a x$ and $y = m x$ is $\frac{a^{2}}{3}$ , then the value of $m$ is:

$2$
$- 2$
$\frac{1}{2}$
None of these

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Ans. 2

Explanation:

Given, $y^{2} = 4 a x$ $\dots (i)$

$y = m x$ $\dots (i i)$

Substitute (ii) in (i):

${(m x)}^{2} = 4 a x$

which gives $m^{2} x^{2} - 4 a x = 0$

$x [m^{2} x - 4 a] = 0$

i.e., $x = 0, x = \frac{4 a}{m^{2}}$

From (ii), equation $x = 0$ , $y = 0$

and $x = \frac{4 a}{m^{2}}$

$\Rightarrow$ $y = m \times (\frac{4 a}{m^{2}})$

$y = \frac{4 a}{m}$

$∴$ $Area = \int_{0}^{\frac{4 a}{m^{2}}} y d x - \int_{0}^{\frac{4 a}{m^{2}}} m x d x$

$= \int_{0}^{\frac{4 a}{m^{2}}} (\sqrt{4 a x} - m x) d x$

$= 2 \sqrt{a} {[\frac{x^{3 / 2}}{3 / 2}]}_{0}^{\frac{4 a}{m^{2}}} - m {[\frac{x^{2}}{2}]}_{0}^{\frac{4 a}{m^{2}}}$

$= \frac{2 \sqrt{a} \times 2}{3} [{(\frac{4 a}{m^{2}})}^{3 / 2}] - \frac{m}{2} {[\frac{4 a}{m^{2}}]}^{2} - (0) - (0)$

$= \frac{4 \sqrt{a}}{3} [\frac{8 a^{3 / 2}}{m^{3}}] - \frac{16 a^{2}}{2 m^{3}}$

$= \frac{32 a^{2}}{3 m^{3}} - \frac{16 a^{2}}{2 m^{3}}$

Now it is given

$\frac{32 a^{2}}{3 m^{3}} - \frac{16 a^{2}}{2 m^{3}} = \frac{a^{2}}{3}$

$\Rightarrow$ $\frac{64 a^{2} - 48 a^{2}}{6 m^{3}} = \frac{a^{2}}{3}$

$\Rightarrow$ $\frac{16 a^{2}}{6 m^{3}} = \frac{a^{2}}{3}$

$\Rightarrow$ $\frac{16}{6} \times 3 = m^{3}$

$\Rightarrow$ $m^{3} = 8$

$∴$ $m = 2$

Q 42 :

The area of the region bounded by the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ is:

$20 π$ sq. units
$20 π^{2}$ sq. units
$16 π^{2}$ sq. units
$25 π$ sq. units

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Ans. $20 π$ sq. units

Explanation:

We have

$\frac{x^{2}}{5^{2}} + \frac{y^{2}}{4^{2}} = 1$

Here, $a = \pm 5$ and $b = \pm 4$

and $\frac{y^{2}}{4^{2}} = 1 - \frac{x^{2}}{5^{2}}$

$\Rightarrow$ $y^{2} = 16 (1 - \frac{x^{2}}{25})$

$\Rightarrow$ $y = \sqrt{\frac{16}{25} (25 - x^{2})}$

$\Rightarrow$ $y = \frac{4}{5} \sqrt{5^{2} - x^{2}}$

Therefore, the area enclosed by the ellipse is

$A = 2 \cdot \frac{4}{5} \int_{- 5}^{5} \sqrt{5^{2} - x^{2}} d x$

$= 2 \cdot \frac{8}{5} \int_{0}^{5} \sqrt{5^{2} - x^{2}} d x$

$= 2 \cdot \frac{8}{5} {[\frac{x}{2} \sqrt{5^{2} - x^{2}} + \frac{5^{2}}{2} \sin^{- 1} (\frac{x}{5})]}_{0}^{5}$

$= 2 \cdot \frac{8}{5} [\frac{5}{2} \sqrt{5^{2} - 5^{2}} + \frac{5^{2}}{2} \sin^{- 1} (\frac{5}{5}) - 0 - \frac{25}{2} \cdot 0]$

$= 2 \cdot \frac{8}{5} [\frac{25}{2} \cdot \frac{π}{2}]$

$= \frac{16}{5} \cdot \frac{25 π}{4}$

$= 20 π sq. units.$

Q 43 :

Area of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is:

$π a b$ sq. units
$\frac{1}{2} π a b$ sq. units
$\frac{1}{4} π a b$ sq. units
None of the above

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Ans. $π a b$ sq. units

Explanation:

Since the given equation contains only even powers of $x$ and only even powers of $y$ , the curve is symmetrical about the $Y$ -axis as well as the $X$ -axis.

$∴$ Whole area of the given ellipse

$= 4 (Area of BCO)$

$= 4 \int_{0}^{a} y d x$

$= 4 \int_{0}^{a} \frac{b}{a} \sqrt{a^{2} - x^{2}} d x$

$= 4 a b \int_{0}^{π / 2} (\frac{1 + \cos 2 θ}{2}) d θ$ [Putting $x = a \sin θ$ ]

$= 2 a b (\int_{0}^{π / 2} d θ + \int_{0}^{π / 2} \cos 2 θ d θ)$

$= 2 a b [{[θ]}_{0}^{π / 2} - {[\frac{\sin 2 θ}{2}]}_{0}^{π / 2}]$

$= 2 a b [\frac{π}{2} - 0 - (0 - 0)]$

$= π a b sq. units.$

Q 44 :

The area enclosed by the curve $y = - x^{2}$ and the straight line $x + y + 2 = 0$ is:

$\frac{9}{2}$ sq. units
$\frac{7}{2}$ sq. units
$\frac{3}{2}$ sq. units
5 sq. units

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Ans. $\frac{9}{2}$ sq. units

Explanation:

We have $y = - x^{2}$ and $x + y + 2 = 0$

$\Rightarrow$ $- x - 2 = - x^{2}$

$\Rightarrow$ $x^{2} - x - 2 = 0$

$\Rightarrow$ $x^{2} + x - 2 x - 2 = 0$

$\Rightarrow$ $x (x + 1) - 2 (x + 1) = 0$

$\Rightarrow$ $(x - 2) (x + 1) = 0$

$\Rightarrow$ $x = 2, - 1$

Hence, the area of the shaded region is

$A =$ $| \int_{- 1}^{2} (- x - 2 + x^{2}) d x |$

$=$ $| \int_{- 1}^{2} (x^{2} - x - 2) d x |$

$=$ $| {[\frac{x^{3}}{3} - \frac{x^{2}}{2} - 2 x]}_{- 1}^{2} |$

$=$ $| [\frac{8}{3} - \frac{4}{2} - 4 + \frac{1}{3} + \frac{1}{2} - 2] |$

$=$ $| \frac{16 - 12 - 24 + 2 + 3 - 12}{6} |$

$=$ $| - \frac{27}{6} |$

$= \frac{9}{2} sq. units.$

Q 45 :

The area of the region enclosed by the parabola $x^{2} = y$ and the line $y = x + 2$ is:

$3$ sq. units
$5$ sq. units
$\frac{9}{2}$ sq. units
$\frac{7}{2}$ sq. units

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Ans. $\frac{9}{2}$ sq. units

Explanation:

We have, $x^{2} = y$ and $y = x + 2$

$\Rightarrow$ $x^{2} = x + 2$

$\Rightarrow$ $x^{2} - x - 2 = 0$

$\Rightarrow$ $x^{2} - 2 x + x - 2 = 0$

$\Rightarrow x (x - 2) + 1 (x - 2) = 0$

$\Rightarrow$ $(x + 1) (x - 2) = 0$

$x = - 1, 2$

$∴$ Required area of the shaded region,

$= \int_{- 1}^{2} (x + 2 - x^{2}) d x$

$= {[\frac{x^{2}}{2} + 2 x - \frac{x^{3}}{3}]}_{- 1}^{2}$

$= (\frac{4}{2} + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3})$

$= 6 + \frac{3}{2} - \frac{9}{3}$

$= \frac{36 + 9 - 18}{6}$

$= \frac{27}{6}$

$= \frac{9}{2} sq. units$

Q 46 :

The area between $x = y^{2}$ and $x = 4$ is divided into two equal parts by the line $x = a$ , the value of $a$ is:

$4$
$\sqrt[3]{16}$
$\sqrt[3]{32}$
None of these

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Ans. $\sqrt[3]{16}$

Explanation :

Given, $x = y^{2}$ is a parabola symmetric to the positive $X$ -axis with vertex (0, 0).

We have given that,

$\int_{0}^{a} \sqrt{x} d x = \int_{a}^{4} \sqrt{x} d x$

$\Rightarrow$ ${[\frac{x^{3 / 2}}{\frac{3}{2}}]}_{0}^{a} = {[\frac{x^{3 / 2}}{\frac{3}{2}}]}_{a}^{4}$

$\Rightarrow$ $\frac{2}{3} a^{3 / 2} = \frac{2}{3} (4^{3 / 2} - a^{3 / 2})$

$\Rightarrow$ $\frac{2}{3} a^{3 / 2} + \frac{2}{3} a^{3 / 2} = \frac{2}{3} \times 8$

$\Rightarrow$ $\frac{4}{3} a^{3 / 2} = \frac{16}{3}$

$\Rightarrow$ $a \sqrt{a} = \frac{16}{3} \times \frac{3}{4}$

$\Rightarrow$ $a^{3 / 2} = 4$

$\Rightarrow$ $a = 4^{2 / 3} = {(2^{2})}^{2 / 3}$

$\Rightarrow$ $a = 2^{4 / 3} = \sqrt[3]{16}$

Q 47 :

The area bounded by $y = 2 - x^{2}$ and $x + y = 0$ is:

$\frac{7}{2}$ sq. units
$\frac{9}{2}$ sq. units
9 sq. units
None of these

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Ans. $\frac{9}{2}$ sq. units

Explanation :

$y = 2 - x^{2}$ and $x + y = 0$

$\Rightarrow$ $x^{2} - x - 2 = 0$

$\Rightarrow$ $x^{2} - 2 x + x - 2 = 0$

$\Rightarrow$ $(x - 2) (x + 1) = 0$

$\Rightarrow$ $x = 2, - 1$

$x = 2$ then $y = - 2$

$x = - 1$ then $y = 1$

Required area,

$A = \int_{- 1}^{2} (2 - x^{2}) d x - \int_{- 1}^{2} (- x) d x$

$= {[2 x - \frac{x^{3}}{3} + \frac{x^{2}}{2}]}_{- 1}^{2}$

$= 4 - \frac{8}{3} + \frac{4}{2} + 2 + \frac{1}{3} + \frac{1}{2}$

$= 6 + 2 - 3 - \frac{1}{2} = \frac{9}{2} sq. units.$

Q 48 :

The area between $X$ -axis and the curve $y = \cos x$ when $0 \leq x \leq 2 π$ is:

0 sq. unit
2 sq. unit
3 sq. unit
4 sq. unit

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Ans. 4 sq. unit

Explanation:

$y = \cos x,$ $0 \leq x \leq 2 π$

$\cos x \geq 0 when x \in [0, \frac{π}{2}]$

$\cos x \leq 0 when x \in [\frac{π}{2}, \frac{3 π}{2}]$

$\cos x \geq 0 when x \in [\frac{3 π}{2}, 2 π]$

$∴$ $Area = \int_{0}^{π / 2} \cos x d x + \int_{π / 2}^{3 π / 2} \cos x d x + \int_{3 π / 2}^{2 π} \cos x d x$

$= 1 + 2 + 1$

$= 4 sq. units$

Q 49 :

The area bounded by the curve $y = \sin x$ between the ordinates $x = 0$ and $x = π$ and the $X$ -axis is:

2 sq. units
4 sq. units
3 sq. units
1 sq. units

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Ans. 2 sq. units

Explanation :

$A = \int_{0}^{π} \sin x d x = {[- \cos x]}_{0}^{π}$

$= 1 + 1 = 2 sq. units$

Q 50 :

The area bounded by the curve $y = x^{4} - 2 x^{3} + x^{2} + 3$ with the $X$ -axis and ordinate corresponding to the minimum of $y$ is:

$1$ sq. units
$\frac{91}{30}$ sq. units
$\frac{30}{9}$ sq. units
$4$ sq. units

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Ans. $\frac{91}{30}$ sq. units

Explanation:

$y = x^{4} - 2 x^{3} + x^{2} + 3$

$\frac{d y}{d x} = 4 x^{3} - 6 x^{2} + 2 x$

$\frac{d y}{d x} = 0$

$\Rightarrow$ $4 x^{3} - 6 x^{2} + 2 x = 0$

$\Rightarrow$ $x (4 x^{2} - 6 x + 2) = 0$

$\Rightarrow$ $2 x (2 x^{2} - 3 x + 1) = 0$

$\Rightarrow$ $x (2 x^{2} - 2 x - x + 1) = 0$

$\Rightarrow$ $x (2 x - 1) (x - 1) = 0$

$x = 0, \frac{1}{2}, 1$

$\frac{d^{2} y}{d x^{2}} = 12 x^{2} - 12 x + 2$

${(\frac{d^{2} y}{d x^{2}})}_{x = 0} = 2 > 0,$

${(\frac{d^{2} y}{d x^{2}})}_{x = \frac{1}{2}} = - 1 < 0,$

${(\frac{d^{2} y}{d x^{2}})}_{x = 1} = 2 > 0$

Hence, minimum occurs at $x = 0$ and $x = 1$

$∴$ $Area = \int_{0}^{1} (x^{4} - 2 x^{3} + x^{2} + 3) d x$

$= {[\frac{x^{5}}{5} - \frac{2 x^{4}}{4} + \frac{x^{3}}{3} + 3 x]}_{0}^{1}$

$= \frac{91}{30} sq. units$

Q 51 :

The area enclosed between $y^{2} = 4 x,$ $x = 1,$ $x = 4$ in the first quadrant is:

$\frac{28}{3}$ sq. unit
$\frac{27}{2}$ sq. unit
$\frac{25}{2}$ sq. unit
$\frac{27}{5}$ sq. unit

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Ans. $\frac{28}{3}$ sq. unit

Explanation :

We have,

$y^{2} = 4 x$

$y = 2 \sqrt{x}$

So, the area enclosed is given by

$Area = \int_{1}^{4} y d x$

$= \int_{1}^{4} 2 \sqrt{x} d x$

$= 2 \int_{1}^{4} x^{1 / 2} d x$

$= 2 {[\frac{2 x^{3 / 2}}{3}]}_{1}^{4}$

$= \frac{4}{3} [4^{3 / 2} - 1]$

$= \frac{4}{3} [8 - 1]$

$= \frac{4}{3} \times 7 = \frac{28}{3} sq. units$

Q 52 :

Area of the region bounded by the curve $y = \cos x$ and the $X$ -axis between $x = 0$ and $x = π$ is:

2 sq. units
3 sq. units
4 sq. units
1 sq. unit

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Ans. 2 sq. units

Explanation:

$Area = \int_{0}^{π / 2} y d x - \int_{π / 2}^{π} y d x$

$= \int_{0}^{π / 2} \cos x d x - \int_{π / 2}^{π} \cos x d x$

$= {[\sin x]}_{0}^{π / 2} - {[\sin x]}_{π / 2}^{π}$

$= (1 - 0) - (0 - 1) = 2 sq. units.$

Q 53 :

The area enclosed between the curve $y = x^{2} + 2$ and the X-axis between $x = 0$ and $x = 3$ is:

14 sq. units
15 sq. units
16 sq. units
18 sq. units

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Ans. 15 sq. units

Explanation:

$Area of LCBA = \int_{0}^{3} y d x$

$= \int_{0}^{3} (x^{2} + 2) d x$

$= {[\frac{x^{3}}{3} + 2 x]}_{0}^{3}$

$= \frac{27}{3} + 6 - 0$

$= 9 + 6 = 15 sq. units.$

Q 54 :

Let $' q'$ be the maximum integral value of $p$ in [0, 10] for which the roots of the equation $x^{2} + p x + \frac{5 p}{4} = 0$ are rational then the area of region ${(x, y) : 0 \leq y \leq {(x - q)}^{2}, 0 \leq x \leq q}$ is (in square units)

$243$
$25$
$\frac{125}{3}$
$164$

1

2

3

4

(1)

$D = p^{2} - 5 p . If p = 9, then D is perfect square$

$Area = \int_{0}^{9} {(x - 9)}^{2} d x = 243$

Q 55 :

The smaller area (in sq. units) included between the curves $\sqrt{x} + \sqrt{| y |} = 1$ and $| x |$ $+$ $| y |$ $= 1$ is

$\frac{1}{3}$
$\frac{4}{3}$
$\frac{2}{3}$
$\frac{5}{3}$

1

2

3

4

(3)

$\sqrt{x} + \sqrt{| y |} = 1 .$

$Above curve is symmetric about x -axis$ $\sqrt{| y |} = 1 - \sqrt{x} and \sqrt{x} = 1 - \sqrt{| y |}$

[IMAGE 138]

$\Rightarrow x > 0, y = 0$

$\sqrt{y} = 1 - \sqrt{x}$

$\frac{1}{2 \sqrt{y}} \frac{d y}{d x} = - \frac{1}{2 \sqrt{x}}$

$\frac{d y}{d x} = - \sqrt{\frac{y}{x}} < 0$

$Function is decreasing$

$Required area = 2 \int_{0}^{1} (2 \sqrt{x} - 2 x) d x = \frac{2}{3}$

Q 56 :

The area of the region between the curves $y = \sqrt{\frac{1 + \sin x}{\cos x}}$ and $y = \sqrt{\frac{1 - \sin x}{\cos x}}$ bounded by the lines $x = 0$ and $x = \frac{π}{4}$ is

$\int_{\sqrt{2} + 1}^{\sqrt{2} - 1} \frac{t}{(1 + t^{2}) \sqrt{1 - t^{2}}} d t$
$\int_{0}^{\sqrt{2} - 1} \frac{4 t}{(1 + t^{2}) \sqrt{1 - t^{2}}} d t$
$\int_{0}^{\sqrt{2} + 1} \frac{4 t}{(1 + t^{2}) \sqrt{1 - t^{2}}} d t$
$\int_{0}^{\sqrt{2} + 1} \frac{t}{(1 + t^{2}) \sqrt{1 - t^{2}}} d t$

1

2

3

4

(2)

$Area = \int_{0}^{\frac{π}{4}} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{\sqrt{\cos x}} d x$

$= \int_{0}^{\frac{π}{4}} \frac{2 \sin \frac{x}{2} d x}{\sqrt{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}}}$

$By putting \tan \frac{x}{2} = t, integral will become$

$\int_{0}^{\sqrt{2} - 1} \frac{4 t}{(1 + t^{2}) \sqrt{1 - t^{2}}} d t$

Q 57 :

Consider the given statements

Statement I: If $f (x)$ is bounded for $x \in [a, b]$ , then area bounded by curve $y = f (x)$ , x-axis, $x = a$ and $x = b$ is $\int_{a}^{b} f (x) d x$

Statement II: If $f (x)$ is bounded and differentiable for $x \in [a, b]$ , then the area between $y = f (x)$ and $y = f^{- 1} (x)$ is equal to double the value of $\int_{a}^{b}$ $| f (x) - x |$ $d x$

Only Statement I is true
Only Statement II is true
Both Statement I and II are True
Both Statement I and II are false

1

2

3

4

(4)

$Use definition of Area.$ $S_{1} (F),$ $S_{2} (F)$

Q 58 :

Let $f (x) = Max {x^{2}, {(1 - x)}^{2}, 2 x (1 - x)}$ where $0 \leq x \leq 1$ , if the area of the region bounded by the curves $y = f (x)$ , $x$ -axis, $x = 0$ and $x = 1$ is $\frac{p}{q}$ then $p + q =$ ____ (H.C.F of $p, q$ is 1)

(44)

From fig it clear that $f (x) =$ ${\begin{matrix} {(1 - x)}^{2}, & 0 \leq x \leq \frac{1}{3} \\ 2 x (1 - x), & \frac{1}{3} < x \leq \frac{2}{3} \\ x^{2}, & \frac{2}{3} < x \leq 1 \end{matrix}$

The required area $A = \int_{0}^{1} f (x)$ $= \int_{0}^{\frac{1}{3}} {(1 - x)}^{2} d x + \int_{\frac{1}{3}}^{\frac{2}{3}} 2 x (1 - x) d x + \int_{\frac{2}{3}}^{1} x^{2} d x$

$= {[- \frac{1}{3} {(1 - x)}^{3}]}_{0}^{\frac{1}{3}} + {[x^{2} - \frac{2 x^{3}}{3}]}_{\frac{1}{3}}^{\frac{2}{3}} +$ ${[\frac{x^{3}}{3}]}_{\frac{2}{3}}^{1}$ $= \frac{17}{27}$

So, $\frac{p}{q} = \frac{17}{27}$

Hence, $p + q = 17 + 27 = 44$

Q 59 :

The area of the region $S = {(x, y) : y^{2} \leq 8 x, y \geq \sqrt{2 x}, x \geq 1}$ is

$\frac{13 \sqrt{2}}{6}$
$\frac{11 \sqrt{2}}{6}$
$\frac{5 \sqrt{2}}{6}$
$\frac{19 \sqrt{2}}{6}$

1

2

3

4

(2)

$Area: \int_{1}^{4} (2 \sqrt{2} \sqrt{x} - \sqrt{2 x}) d x$

Q 60 :

Let $f (x) = \max {x^{2}, {(1 - x)}^{2}, 2 x (1 - x)}$ where $0 \leq x \leq 1$ , if the Area of the region bounded by the curves $y = f (x)$ , $x$ -axis, $x = 0$ and $x = 1$ is $\frac{p}{q}$ ; (where $p, q$ are coprime numbers) then $p + q =$ ____.

30
40
44
72

1

2

3

4

(3)

$The required Area A = \int_{0}^{1} f (x) d x$

[IMAGE 236]

$= \int_{0}^{1 / 3} {(1 - x)}^{2} d x + \int_{1 / 3}^{2 / 3} 2 x (1 - x) d x + \int_{2 / 3}^{1} x^{2} d x$

$= {[- \frac{1}{3} {(1 - x)}^{3}]}_{0}^{1 / 3} + {[x^{2} - \frac{2 x^{3}}{3}]}_{1 / 3}^{2 / 3} +$ ${[\frac{x^{3}}{3}]}_{\frac{2}{3}}^{1}$

$= \frac{17}{27}$

Topic Question Set

Q 31 :

The area of the region bounded by parabola $y^{2} = x$ and the straight line $2 y = x$ is:

Q 32 :

Area bounded by the parabola $y = 4 x^{2},$ Y-axis and the lines $y = 1, y = 4$ is:

Q 33 :

The area bounded by the parabola $y^{2} = 4 a x,$ its axis and two ordinates $x = 4, x = 9$ is:

Q 34 :

Area bounded by the parabola $y^{2} = 4 a x$ and its latus-rectum is:

Q 35 :

The area between the curve $y^{2} = 4 a x$ , the $x$ -axis and the lines $x = 0$ and $x = a$ is:

Q 36 :

The area bounded by the curve $y = x^{2}$ , the $Y$ -axis and the $X$ -axis and $x = 3$ is:

Q 37 :

The area of the regions bounded by the curve $y = x^{2}$ , the $X$ -axis and the lines $x = 1$ and $x = 3$ is:

Q 38 :

The area of the regions bounded by the curve $y^{2} = 4 x$ , the $X$ -axis and the lines $x = 1$ , $x = 4, y \geq 0$ is:

Q 39 :

The area of the given bounded by the curves $x^{2} = 16 y$ , $y = 1$ , $y = 4$ and the $Y$ -axis, lying in the first quadrant:

Q 40 :

The area of the region bounded by the curve $x^{2} = 4 y$ and the straight line $x = 4 y - 2$ is:

Q 41 :

If the area bounded by the curve $y^{2} = 4 a x$ and $y = m x$ is $\frac{a^{2}}{3}$ , then the value of $m$ is:

Q 42 :

The area of the region bounded by the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ is:

Q 43 :

Area of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is:

Q 44 :

The area enclosed by the curve $y = - x^{2}$ and the straight line $x + y + 2 = 0$ is:

Q 45 :

The area of the region enclosed by the parabola $x^{2} = y$ and the line $y = x + 2$ is:

Q 46 :

The area between $x = y^{2}$ and $x = 4$ is divided into two equal parts by the line $x = a$ , the value of $a$ is:

Q 47 :

The area bounded by $y = 2 - x^{2}$ and $x + y = 0$ is:

Q 48 :

The area between $X$ -axis and the curve $y = \cos x$ when $0 \leq x \leq 2 π$ is:

Q 49 :

The area bounded by the curve $y = \sin x$ between the ordinates $x = 0$ and $x = π$ and the $X$ -axis is:

Q 50 :

The area bounded by the curve $y = x^{4} - 2 x^{3} + x^{2} + 3$ with the $X$ -axis and ordinate corresponding to the minimum of $y$ is:

Q 51 :

The area enclosed between $y^{2} = 4 x,$ $x = 1,$ $x = 4$ in the first quadrant is:

Q 52 :

Area of the region bounded by the curve $y = \cos x$ and the $X$ -axis between $x = 0$ and $x = π$ is:

Q 53 :

The area enclosed between the curve $y = x^{2} + 2$ and the X-axis between $x = 0$ and $x = 3$ is:

Q 54 :

Let $' q'$ be the maximum integral value of $p$ in [0, 10] for which the roots of the equation $x^{2} + p x + \frac{5 p}{4} = 0$ are rational then the area of region ${(x, y) : 0 \leq y \leq {(x - q)}^{2}, 0 \leq x \leq q}$ is (in square units)

Q 55 :

The smaller area (in sq. units) included between the curves $\sqrt{x} + \sqrt{| y |} = 1$ and $| x |$ $+$ $| y |$ $= 1$ is

Q 56 :

The area of the region between the curves $y = \sqrt{\frac{1 + \sin x}{\cos x}}$ and $y = \sqrt{\frac{1 - \sin x}{\cos x}}$ bounded by the lines $x = 0$ and $x = \frac{π}{4}$ is

Q 58 :

Let $f (x) = Max {x^{2}, {(1 - x)}^{2}, 2 x (1 - x)}$ where $0 \leq x \leq 1$ , if the area of the region bounded by the curves $y = f (x)$ , $x$ -axis, $x = 0$ and $x = 1$ is $\frac{p}{q}$ then $p + q =$ ____ (H.C.F of $p, q$ is 1)

Q 59 :

The area of the region $S = {(x, y) : y^{2} \leq 8 x, y \geq \sqrt{2 x}, x \geq 1}$ is

Q 60 :

Let $f (x) = \max {x^{2}, {(1 - x)}^{2}, 2 x (1 - x)}$ where $0 \leq x \leq 1$ , if the Area of the region bounded by the curves $y = f (x)$ , $x$ -axis, $x = 0$ and $x = 1$ is $\frac{p}{q}$ ; (where $p, q$ are coprime numbers) then $p + q =$ ____.

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Topic Question Set

Q 31 : The area of the region bounded by parabola y2=x and the straight line 2y=x is:

Q 32 : Area bounded by the parabola y=4x2, Y-axis and the lines y=1, y=4 is:

Q 33 : The area bounded by the parabola y2=4ax, its axis and two ordinates x=4, x=9 is:

Q 34 : Area bounded by the parabola y2=4ax and its latus-rectum is:

Q 35 : The area between the curve y2=4ax, the x-axis and the lines x=0 and x=a is:

Q 36 : The area bounded by the curve y=x2, the Y-axis and the X-axis and x=3 is:

Q 37 : The area of the regions bounded by the curve y=x2, the X-axis and the lines x=1 and x=3 is:

Q 38 : The area of the regions bounded by the curve y2=4x, the X-axis and the lines x=1, x=4, y≥0 is:

Q 39 : The area of the given bounded by the curves x2=16y, y=1, y=4 and the Y-axis, lying in the first quadrant:

Q 40 : The area of the region bounded by the curve x2=4y and the straight line x=4y-2 is:

Q 41 : If the area bounded by the curve y2=4ax and y=mx is a23, then the value of m is:

Q 42 : The area of the region bounded by the ellipse x225+y216=1 is:

Q 43 : Area of the ellipse x2a2+y2b2=1 is:

Q 44 : The area enclosed by the curve y=-x2 and the straight line x+y+2=0 is:

Q 45 : The area of the region enclosed by the parabola x2=y and the line y=x+2 is:

Q 46 : The area between x=y2 and x=4 is divided into two equal parts by the line x=a, the value of a is:

Q 47 : The area bounded by y=2-x2 and x+y=0 is:

Q 48 : The area between X-axis and the curve y=cosx when 0≤x≤2π is:

Q 49 : The area bounded by the curve y=sinx between the ordinates x=0 and x=π and the X-axis is:

Q 50 : The area bounded by the curve y=x4-2x3+x2+3 with the X-axis and ordinate corresponding to the minimum of y is:

Q 51 : The area enclosed between y2=4x, x=1, x=4 in the first quadrant is:

Q 52 : Area of the region bounded by the curve y=cosx and the X-axis between x=0 and x=π is:

Q 53 : The area enclosed between the curve y=x2+2 and the X-axis between x=0 and x=3 is:

Q 54 : Let 'q' be the maximum integral value of p in [0, 10] for which the roots of the equation x2+px+5p4=0 are rational then the area of region {(x,y):0≤y≤(x-q)2, 0≤x≤q} is (in square units)

Q 55 : The smaller area (in sq. units) included between the curves x+|y|=1 and |x|+|y|=1 is

Q 56 : The area of the region between the curves y=1+sinxcosx and y=1-sinxcosx bounded by the lines x=0 and x=π4 is

Q 58 : Let f(x)=Max{x2,(1-x)2,2x(1-x)} where 0≤x≤1, if the area of the region bounded by the curves y=f(x), x-axis, x=0 and x=1 is pq then p+q=____ (H.C.F of p,q is 1)

Q 59 : The area of the region S={(x,y):y2≤8x, y≥2x, x≥1} is

Q 60 : Let f(x)=max{x2,(1-x)2,2x(1-x)} where 0≤x≤1, if the Area of the region bounded by the curves y=f(x), x-axis, x=0 and x=1 is pq; (where p,q are coprime numbers) then p+q=____.

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Q 31 :

The area of the region bounded by parabola $y^{2} = x$ and the straight line $2 y = x$ is:

Q 32 :

Area bounded by the parabola $y = 4 x^{2},$ Y-axis and the lines $y = 1, y = 4$ is:

Q 33 :

The area bounded by the parabola $y^{2} = 4 a x,$ its axis and two ordinates $x = 4, x = 9$ is:

Q 34 :

Area bounded by the parabola $y^{2} = 4 a x$ and its latus-rectum is:

Q 35 :

The area between the curve $y^{2} = 4 a x$ , the $x$ -axis and the lines $x = 0$ and $x = a$ is:

Q 36 :

The area bounded by the curve $y = x^{2}$ , the $Y$ -axis and the $X$ -axis and $x = 3$ is:

Q 37 :

The area of the regions bounded by the curve $y = x^{2}$ , the $X$ -axis and the lines $x = 1$ and $x = 3$ is:

Q 38 :

The area of the regions bounded by the curve $y^{2} = 4 x$ , the $X$ -axis and the lines $x = 1$ , $x = 4, y \geq 0$ is:

Q 39 :

The area of the given bounded by the curves $x^{2} = 16 y$ , $y = 1$ , $y = 4$ and the $Y$ -axis, lying in the first quadrant:

Q 40 :

The area of the region bounded by the curve $x^{2} = 4 y$ and the straight line $x = 4 y - 2$ is:

Q 41 :

If the area bounded by the curve $y^{2} = 4 a x$ and $y = m x$ is $\frac{a^{2}}{3}$ , then the value of $m$ is:

Q 42 :

The area of the region bounded by the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ is:

Q 43 :

Area of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is:

Q 44 :

The area enclosed by the curve $y = - x^{2}$ and the straight line $x + y + 2 = 0$ is:

Q 45 :

The area of the region enclosed by the parabola $x^{2} = y$ and the line $y = x + 2$ is:

Q 46 :

The area between $x = y^{2}$ and $x = 4$ is divided into two equal parts by the line $x = a$ , the value of $a$ is:

Q 47 :

The area bounded by $y = 2 - x^{2}$ and $x + y = 0$ is:

Q 48 :

The area between $X$ -axis and the curve $y = \cos x$ when $0 \leq x \leq 2 π$ is:

Q 49 :

The area bounded by the curve $y = \sin x$ between the ordinates $x = 0$ and $x = π$ and the $X$ -axis is:

Q 50 :

The area bounded by the curve $y = x^{4} - 2 x^{3} + x^{2} + 3$ with the $X$ -axis and ordinate corresponding to the minimum of $y$ is:

Q 51 :

The area enclosed between $y^{2} = 4 x,$ $x = 1,$ $x = 4$ in the first quadrant is:

Q 52 :

Area of the region bounded by the curve $y = \cos x$ and the $X$ -axis between $x = 0$ and $x = π$ is:

Q 53 :

The area enclosed between the curve $y = x^{2} + 2$ and the X-axis between $x = 0$ and $x = 3$ is:

Q 54 :

Let $' q'$ be the maximum integral value of $p$ in [0, 10] for which the roots of the equation $x^{2} + p x + \frac{5 p}{4} = 0$ are rational then the area of region ${(x, y) : 0 \leq y \leq {(x - q)}^{2}, 0 \leq x \leq q}$ is (in square units)

Q 55 :

The smaller area (in sq. units) included between the curves $\sqrt{x} + \sqrt{| y |} = 1$ and $| x |$ $+$ $| y |$ $= 1$ is

Q 56 :

The area of the region between the curves $y = \sqrt{\frac{1 + \sin x}{\cos x}}$ and $y = \sqrt{\frac{1 - \sin x}{\cos x}}$ bounded by the lines $x = 0$ and $x = \frac{π}{4}$ is

Q 58 :

Let $f (x) = Max {x^{2}, {(1 - x)}^{2}, 2 x (1 - x)}$ where $0 \leq x \leq 1$ , if the area of the region bounded by the curves $y = f (x)$ , $x$ -axis, $x = 0$ and $x = 1$ is $\frac{p}{q}$ then $p + q =$ ____ (H.C.F of $p, q$ is 1)

Q 59 :

The area of the region $S = {(x, y) : y^{2} \leq 8 x, y \geq \sqrt{2 x}, x \geq 1}$ is

Q 60 :

Let $f (x) = \max {x^{2}, {(1 - x)}^{2}, 2 x (1 - x)}$ where $0 \leq x \leq 1$ , if the Area of the region bounded by the curves $y = f (x)$ , $x$ -axis, $x = 0$ and $x = 1$ is $\frac{p}{q}$ ; (where $p, q$ are coprime numbers) then $p + q =$ ____.