Let the area of the region bounded by the curve lines , and the x-axis be A. Then is equal to ________ . [2026]

Question

Let the area of the region bounded by the curve $y = \max {\sin x, \cos x},$ lines $x = 0, x = \frac{3 π}{2}$ , and the $x$ -axis be A. Then $A + A^{2}$ is equal to ________ . [2026]

Smart Achievers · Accepted Answer

(12)

$A = \int_{0}^{π / 4} \cos x d x + \int_{π / 4}^{π} \sin x d x + \int_{π}^{5 π / 4} - \sin x d x + \int_{5 π / 4}^{3 π / 2} - \cos x d x$

$A = {(\sin x)}_{0}^{π / 4} + {(\cos x)}_{π / 4}^{π} + {(\cos x)}_{π}^{5 π / 4} + (\sin x)_{5 π / 4}^{3 π / 2}$

$A = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + 1 + 1 - \frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}} = 3$

$A^{2} + A = 12$

Solution

Q.
Let the area of the region bounded by the curve $y = \max {\sin x, \cos x},$ lines $x = 0, x = \frac{3 π}{2}$ , and the $x$ -axis be A. Then $A + A^{2}$ is equal to ________ . [2026]

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Solution

Q. Let the area of the region bounded by the curve y=max{sinx, cosx}, lines x=0, x=3π2, and the x-axis be A. Then A+A2 is equal to ________ . [2026]

Ans. (12) A=∫0π/4cosx dx+∫π/4πsinx dx+∫π5π/4-sinx dx+∫5π/43π/2-cosx dx A=(sinx)0π/4+(cosx)π/4π+(cosx)π5π/4+(sinx)5π/43π/2 A=12+12+1+1-12+1-12=3 A2+A=12

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Q.
Let the area of the region bounded by the curve $y = \max {\sin x, \cos x},$ lines $x = 0, x = \frac{3 π}{2}$ , and the $x$ -axis be A. Then $A + A^{2}$ is equal to ________ . [2026]