Application of Integrals10 questions7 PYQ

Application of Integrals — JEE Maths Practice Questions & Solutions

10 questions on Application of Integrals with full step-by-step solutions, including past-year (PYQ) problems. Free to practice.

hardPYQ · JEE Main 2026
Let ee be the base of natural logarithm and let f:{1,2,3,4}→{1,e,e2,e3}f:\{1,2,3,4\}\to\{1,e,e^2,e^3\} and g:{1,e,e2,e3}→{1,12,13,14}g:\{1,e,e^2,e^3\}\to\left\{1,\dfrac12,\dfrac13,\dfrac14\right\} be two bijective functions such that ff is strictly decreasing and gg is strictly increasing. If ϕ(x)=[f−1{g−1(12)}]x\phi(x)=\left[f^{-1}\left\{g^{-1}\left(\dfrac12\right)\right\}\right]^x, then the area of the region R={(x,y):x2≤y≤ϕ(x), 0≤x≤1}R=\{(x,y):x^2\le y\le\phi(x),\ 0\le x\le1\} is
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mediumPYQ · JEE Main 2026
The area of the region {(x,y):0≤y≤6−x, y2≥4x−3, x≥0}\{(x,y):0\le y\le6-x,\ y^2\ge4x-3,\ x\ge0\} is
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mediumPYQ · JEE Main 2026
If the area of the region bounded by 16x2−9y2=14416x^2-9y^2=144 and 8x−3y=248x-3y=24 is AA, then 3(A+6log⁡e3)3(A+6\log_e 3) is equal to
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mediumPYQ · JEE Main 2026
The area of the region bounded by the curves x+3y2=0x+3y^2=0 and x+4y2=1x+4y^2=1 is equal to
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mediumPYQ · JEE Main 2026
The area of the region {(x,y):y≤π−∣x∣, y≤∣xsin⁡x∣, y≥0}\{(x,y):y\le\pi-|x|,\ y\le|x\sin x|,\ y\ge0\} is
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mediumPYQ · JEE Main 2026
The area of the region R={(x,y):xy≤27, 1≤y≤x2}R=\{(x,y):xy\le27,\ 1\le y\le x^2\} is
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mediumPYQ · JEE Main 2026
Let the line x=−1x=-1 divide the area of the region {(x,y):1+x2≤y≤3−x}\{(x,y):1+x^2\le y\le3-x\} in the ratio m:nm:n, gcd⁡(m,n)=1\gcd(m,n)=1. Then m+nm+n is
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medium
In the coordinate plane, the region MM consists of all the points (x,y)\left(x,y\right) satisfying the inequalities y≥0y \ge 0, y≤xy \le x and y≤2−xy \le 2-x simultaneously. The region NN, which varies with the parameter tt, consists of all the points (x,y)\left(x,y\right) satisfying the inequalities t≤x≤t+1t \le x \le t+1 and 0≤t≤10 \le t \le 1 simultaneously. The area of M∩NM \cap N is −k1t2+k2t+12-k_1t^2+k_2t+\dfrac12, then k1+k2k_1+k_2 is
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medium
Find the area of the region bounded by the curves y=x2y = x^2, y=∣2−x2∣y = \left|2-x^2\right| and y=2y = 2, which lies to the right of the line x=1x = 1.
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medium
Let PP be a point (not the origin) lying on the parabola x2=yx^2 = y. The normal line to the parabola at PP will intersect the parabola at another point QQ. The minimum possible value for the area bounded by the line PQPQ and the parabola is
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