Original JEE Main and CUET-style questions on integrals, areas and differential equations, with a worked solution and the usual trap for each. Tap an option, or type a numerical answer, to check it.
Learn the chapters first: Class 12, Integrals · Class 12, Application of Integrals · Class 12, Differential Equations.
\(\displaystyle\int_0^{\pi/2} \frac{\sin^5 x}{\sin^5 x + \cos^5 x}\,dx\) equals
Call it \(I\). Replacing \(x\) by \(\tfrac{\pi}{2} - x\) gives \(I = \displaystyle\int_0^{\pi/2}\frac{\cos^5 x}{\cos^5 x + \sin^5 x}\,dx\).
Adding: \(2I = \displaystyle\int_0^{\pi/2} 1\,dx = \frac{\pi}{2}\), so \(I = \dfrac{\pi}{4}\), whatever the power.
\(\displaystyle\int x e^{2x}\,dx\) equals
By parts with \(u = x\), \(dv = e^{2x}\,dx\): \(\dfrac{xe^{2x}}{2} - \displaystyle\int \frac{e^{2x}}{2}\,dx = \frac{xe^{2x}}{2} - \frac{e^{2x}}{4} + C = \frac{e^{2x}}{4}(2x - 1) + C\).
Check by differentiating: \(\dfrac{e^{2x}}{2}(2x - 1) + \dfrac{e^{2x}}{4} \cdot 2 = xe^{2x}\).
The area enclosed by \(y = x^2\) and the line \(y = 4\) is \(A\). Find \(3A\).
The curves meet at \(x = \pm 2\). \(A = \displaystyle\int_{-2}^{2} (4 - x^2)\,dx = 2\left[4x - \frac{x^3}{3}\right]_0^2 = 2\left(8 - \frac83\right) = \frac{32}{3}\).
\(3A = 32\).
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