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

Q 11 :

If $f (x)$ satisfies the relation $f (x) = e^{x} + \int_{0}^{1} (y + x e^{x}) f (y) d y,$ then $e + f (0)$ is equal to _______ . [2026]

(2)

$f (x) = e^{x} + \int_{0}^{1} y f (y) d y + x e^{x} \int_{0}^{1} f (y) d y$

$f (x) = e^{x} + A + B x e^{x}$

$A = \int_{0}^{1} y f (y) d y = \int_{0}^{1} y (A + e^{y} + B y e^{y}) d y$

$A = \frac{A}{2} + 0 - (- 1) + B (e - 1)$

$\frac{A}{2} + B (1 - e) = 1$

$B = \int_{0}^{1} f (y) d y$

$B = \int_{0}^{1} (e^{y} + A + B y e^{y}) d y$

$B = (e - 1) + A + B (0 - (- 1))$

$B = e - 1 + A + B \Rightarrow A = 1 - e$

$f (x) = e^{x} + A + B x e^{x}$

$f (0) = 1 + A = 1 - e + 1 = 2 - e$

$e + f (0) = 2$

Q 12 :

Let $I (x) = \int \frac{3 d x}{(4 x + 6) (\sqrt{4 x^{2} + 8 x + 3})}$ and $I (0) = \frac{\sqrt{3}}{4} + 20 .$ If $I (\frac{1}{2}) = \frac{a \sqrt{2}}{b} + c, where a, b, c \in ℕ, \gcd (a, b) = 1,$ then $a + b + c$ is equal to. [2026]

31
30
29
28

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4

(1)

$Let 4 x + 6 = \frac{1}{t} \Rightarrow x = \frac{\frac{1}{t} - 6}{4}$

$4 d x = - \frac{d t}{t^{2}},$ ${\frac{x + 1 = \frac{1}{t} - 2}{4}$

$\int \frac{3 d x}{(4 x + 6) \sqrt{4 {(x + 1)}^{2} - 1}}$

$= \int \frac{3 (- d t)}{4 t^{2} \cdot \frac{1}{t} \sqrt{4 {(\frac{1 / t - 2}{4})}^{2} - 1}}$

$= - \frac{3}{4} \int \frac{d t}{t \sqrt{\frac{{(1 - 2 t)}^{2}}{4 t^{2}} - 1}}$

$= - \frac{3}{4} \int \frac{d t (2 t)}{t \sqrt{1 - 4} t}$

$= - \frac{3}{2} \int \frac{d t}{\sqrt{1 - 4 t}}$ $= - \frac{3}{2} (\frac{\sqrt{1 - 4 t}}{\frac{1}{2} \times - 4}) + C$

$= \frac{3}{4} \sqrt{1 - 4 t} + C$ $∵$ $t = \frac{1}{4 x + 6}$

$= \frac{3}{4} \sqrt{1 - 4 (\frac{1}{4 x + 6})} + C$

$= \frac{3}{4} \sqrt{\frac{4 x + 6 - 4}{4 x + 6}} + C$

$I (x) = \frac{3}{4} \sqrt{\frac{4 x + 2}{4 x + 6}} + C$

$I (0) = \frac{3}{4} \sqrt{\frac{2}{6}} + C = \frac{\sqrt{3}}{4} + C$

$\Rightarrow C = 20$

$Hence I (x) = \frac{3}{4} \sqrt{\frac{4 x + 2}{4 x + 6}} + 20$

$I (\frac{1}{2}) = \frac{3}{4} \sqrt{\frac{4}{8}} + 20$

$= \frac{3}{4 \sqrt{2}} + 20 = \frac{3 \sqrt{2}}{8} + 20$

$a + b + c = 3 + 8 + 20 = 31$

Q 13 :

If $\int ((\frac{1 - 5 \cos^{2} x}{\sin^{5} x \cos^{2} x})) d x = f (x) + C,$ where C is the constant of integration, then $f (\frac{π}{6}) - f (\frac{π}{4})$ is equal to [2026]

$\frac{4}{\sqrt{3}} (8 - \sqrt{6})$
$\frac{2}{\sqrt{3}} (4 + \sqrt{6})$
$\frac{1}{\sqrt{3}} (26 + \sqrt{3})$
$\frac{1}{\sqrt{3}} (26 - \sqrt{3})$

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(1)

$\int \frac{d x}{\sin^{5} x \cos^{2} x} - 5 \int \frac{d x}{\sin^{5} x}$

$= \int \frac{\sec^{2} x d x}{\sin^{5} x} - 5 \int \frac{d x}{\sin^{5} x}$

$By IBP$

$= \frac{\tan x}{\sin^{5} x} - \int \frac{5}{\sin^{6} x} \cos x \tan x d x - 5 \int \frac{d x}{\sin^{5} x}$

$= \frac{\tan x}{\sin^{5} x} + C$

$f (x) = \frac{\tan x}{\sin^{5} x}$

$f (\frac{π}{6}) - f (\frac{π}{4}) = \frac{2^{5}}{\sqrt{3}} - {(\sqrt{2})}^{5} = 4 \sqrt{2} - \frac{32}{\sqrt{3}}$

$= \frac{32}{\sqrt{3}} - 4 \sqrt{2}$

$= \frac{4}{\sqrt{3}} (8 - \sqrt{6})$

Q 14 :

$\int \frac{6 x^{10} + 4}{x^{3} \sqrt{x^{10} - 3 x^{4} - 1}} d x, x > 0 =$ _____________ (c is constant of integration)

$\frac{\sqrt{x^{10} - 3 x^{4} - 1}}{x^{2}} + c$
$\frac{2 \sqrt{x^{10} - 3 x^{4} - 1}}{x^{2}} + c$
$2 \sqrt{x^{10} - 3 x^{4} - 1} + c$
$\sqrt{x^{10} - 3 x^{4}} + C$

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(2)

$\int \frac{6 x^{10} + 4}{x^{5} \sqrt{x^{6} - 3 - x^{- 4}}} d x = \int \frac{6 x^{5} + 4 x^{- 5}}{\sqrt{x^{6} - 3 - x^{- 4}}} d x$

$x^{6} - 3 - x^{- 4} = t (6 x^{5} + 4 x^{- 5}) d x = d t$

Q 15 :

If $I = \int (x^{2} + 1) ({(x + 1) e^{x})}^{2} d x = A (f {(x))}^{2} + C,$ where C is constant of integration and $f (- 1) = \frac{2}{e},$ then $2 A + f (0)$ is

(2)

$I = \int (x^{2} + 1) ({(x + 1) e^{x})}^{2} d x$

$(x^{2} + 1) e^{x} = t$

$\Rightarrow {(x + 1)}^{2} e^{x} d x = d t$

$I = \frac{1}{2} ({(x^{2} + 1) e^{x})}^{2} + c$

$= \frac{1}{2} (f {(x))}^{2} + c$

$2 A + f (0) = 1 + 1 = 2$

Topic Question Set

Q 11 :

If $f (x)$ satisfies the relation $f (x) = e^{x} + \int_{0}^{1} (y + x e^{x}) f (y) d y,$ then $e + f (0)$ is equal to _______ . [2026]

Q 12 :

Let $I (x) = \int \frac{3 d x}{(4 x + 6) (\sqrt{4 x^{2} + 8 x + 3})}$ and $I (0) = \frac{\sqrt{3}}{4} + 20 .$ If $I (\frac{1}{2}) = \frac{a \sqrt{2}}{b} + c, where a, b, c \in ℕ, \gcd (a, b) = 1,$ then $a + b + c$ is equal to. [2026]

Q 13 :

If $\int ((\frac{1 - 5 \cos^{2} x}{\sin^{5} x \cos^{2} x})) d x = f (x) + C,$ where C is the constant of integration, then $f (\frac{π}{6}) - f (\frac{π}{4})$ is equal to [2026]

Q 14 :

$\int \frac{6 x^{10} + 4}{x^{3} \sqrt{x^{10} - 3 x^{4} - 1}} d x, x > 0 =$ _____________ (c is constant of integration)

Q 15 :

If $I = \int (x^{2} + 1) ({(x + 1) e^{x})}^{2} d x = A (f {(x))}^{2} + C,$ where C is constant of integration and $f (- 1) = \frac{2}{e},$ then $2 A + f (0)$ is

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Topic Question Set

Q 11 : If f(x) satisfies the relation f(x)=ex+∫01(y+xex) f(y) dy, then e+f(0) is equal to _______ . [2026]

Q 12 : Let I(x)=∫3dx(4x+6)(4x2+8x+3) and I(0)=34+20. If I (12)=a2b+c, where a,b,c∈ℕ, gcd(a,b)=1,then a+b+c is equal to. [2026]

Q 13 : If ∫(1-5cos2xsin5x cos2x)dx=f(x)+C, where C is the constant of integration, then f(π6)-f(π4) is equal to [2026]

Q 14 : ∫6x10+4x3x10-3x4-1 dx, x>0= _____________ (c is constant of integration)

Q 15 : If I=∫(x2+1)((x+1)ex)2dx=A(f(x))2+C, where C is constant of integration and f(-1)=2e, then 2A+f(0) is

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Q 11 :

If $f (x)$ satisfies the relation $f (x) = e^{x} + \int_{0}^{1} (y + x e^{x}) f (y) d y,$ then $e + f (0)$ is equal to _______ . [2026]

Q 12 :

Let $I (x) = \int \frac{3 d x}{(4 x + 6) (\sqrt{4 x^{2} + 8 x + 3})}$ and $I (0) = \frac{\sqrt{3}}{4} + 20 .$ If $I (\frac{1}{2}) = \frac{a \sqrt{2}}{b} + c, where a, b, c \in ℕ, \gcd (a, b) = 1,$ then $a + b + c$ is equal to. [2026]

Q 13 :

If $\int ((\frac{1 - 5 \cos^{2} x}{\sin^{5} x \cos^{2} x})) d x = f (x) + C,$ where C is the constant of integration, then $f (\frac{π}{6}) - f (\frac{π}{4})$ is equal to [2026]

Q 14 :

$\int \frac{6 x^{10} + 4}{x^{3} \sqrt{x^{10} - 3 x^{4} - 1}} d x, x > 0 =$ _____________ (c is constant of integration)

Q 15 :

If $I = \int (x^{2} + 1) ({(x + 1) e^{x})}^{2} d x = A (f {(x))}^{2} + C,$ where C is constant of integration and $f (- 1) = \frac{2}{e},$ then $2 A + f (0)$ is