Master Practice Matrix — 40 JEE Solved Problems

1\(\displaystyle\int\frac{x}{\sqrt{1-x^4}}\,dx\) Solution: Let \(t=x^2\), \(dt=2x\,dx\). \(\displaystyle\frac{1}{2}\int\frac{dt}{\sqrt{1-t^2}} = \frac{1}{2}\sin^{-1}t+C = \boxed{\frac{1}{2}\sin^{-1}(x^2)+C}\)

2\(\displaystyle\int\frac{\ln x}{x^2}\,dx\) Solution — IBP: \(u=\ln x\), \(v=x^{-2}\): \(= \ln x\cdot\left(-\dfrac{1}{x}\right) - \displaystyle\int\left(-\frac{1}{x}\right)\cdot\frac{1}{x}\,dx = -\frac{\ln x}{x}+\int\frac{dx}{x^2} = -\frac{\ln x}{x}-\frac{1}{x}+C = \boxed{-\frac{1+\ln x}{x}+C}\)

3\(\displaystyle\int e^x\!\left(\frac{1}{x}-\frac{1}{x^2}\right)dx\) Solution — \(e^x[f+f']\) form with \(f=1/x\): \(f(x)=\dfrac{1}{x}\), \(f'(x)=-\dfrac{1}{x^2}\) ✓. \(\boxed{=\dfrac{e^x}{x}+C}\)

4\(\displaystyle\int\frac{\cos x-\sin x}{\cos x+\sin x}\,dx\) Solution — Numerator = derivative of denominator: Let \(t=\cos x+\sin x\), \(dt=(\cos x-\sin x)\,dx\). \(\displaystyle\int\frac{dt}{t} = \ln|t|+C = \boxed{\ln|\sin x+\cos x|+C}\)

5\(\displaystyle\int\frac{dx}{1+\sin x}\) Solution — Multiply by \(\frac{1-\sin x}{1-\sin x}\): \(\displaystyle\int\frac{1-\sin x}{\cos^2 x}\,dx = \int(\sec^2 x-\sec x\tan x)\,dx = \tan x-\sec x+C\)

6\(\displaystyle\int x^2\sin x\,dx\) Solution — IBP twice (\(u=x^2\)): Round 1: \(x^2(-\cos x)-\displaystyle\int 2x(-\cos x)\,dx = -x^2\cos x+2\int x\cos x\,dx\)
Round 2: \(\displaystyle\int x\cos x\,dx = x\sin x+\cos x\)
\(\boxed{= -x^2\cos x+2x\sin x+2\cos x+C}\)

7\(\displaystyle\int\frac{\sqrt{\tan x}}{\sin x\cos x}\,dx\) Solution — Convert using \(\sin x\cos x=\frac{\tan x}{\sec^2 x}\): \(\displaystyle\int\frac{\sqrt{\tan x}\cdot\sec^2 x}{\tan x}\,dx = \int\frac{\sec^2 x}{\sqrt{\tan x}}\,dx\)
Let \(t=\tan x\), \(dt=\sec^2 x\,dx\): \(\displaystyle\int\frac{dt}{\sqrt{t}} = 2\sqrt{t}+C = \boxed{2\sqrt{\tan x}+C}\)

8\(\displaystyle\int\frac{dx}{x(x^n+1)}\) Solution — Multiply top & bottom by \(x^{n-1}\): \(\displaystyle\int\frac{x^{n-1}}{x^n(x^n+1)}\,dx\). Let \(t=x^n\), \(dt=nx^{n-1}dx\):
\(\dfrac{1}{n}\displaystyle\int\frac{dt}{t(t+1)} = \dfrac{1}{n}\int\left(\frac{1}{t}-\frac{1}{t+1}\right)dt = \dfrac{1}{n}\ln\left|\dfrac{t}{t+1}\right|+C = \boxed{\dfrac{1}{n}\ln\left|\dfrac{x^n}{x^n+1}\right|+C}\)

9\(\displaystyle\int\sqrt{\frac{1-x}{1+x}}\,dx\) Solution — Rationalize: \(\displaystyle\int\frac{1-x}{\sqrt{1-x^2}}\,dx = \int\frac{1}{\sqrt{1-x^2}}\,dx - \int\frac{x}{\sqrt{1-x^2}}\,dx\)
\(= \sin^{-1}x+\sqrt{1-x^2}+C\)

10\(\displaystyle\int\frac{2x+1}{(x^2+4x+1)^{3/2}}\,dx\) Solution — Numerator \(2x+1\) is almost derivative of \(x^2+4x+1\) (which is \(2x+4\)): Write \(2x+1 = (2x+4)-3\). Split:
\(\displaystyle\int\frac{2x+4}{(x^2+4x+1)^{3/2}}\,dx - 3\int\frac{dx}{(x^2+4x+1)^{3/2}}\)
First: \(=-2(x^2+4x+1)^{-1/2}\). Second needs completing square. Final answer involves \(-\frac{2}{\sqrt{x^2+4x+1}}\) plus inverse trig term.

11\(\displaystyle\int\sin^3 x\cos^4 x\,dx\) Solution — Odd power of sin: substitute \(t=\cos x\): \(\sin^3 x = \sin x(1-\cos^2 x)\). Let \(t=\cos x\), \(dt=-\sin x\,dx\):
\(\displaystyle-\int(1-t^2)t^4\,dt = -\int(t^4-t^6)\,dt = -\frac{t^5}{5}+\frac{t^7}{7}+C = \boxed{-\frac{\cos^5 x}{5}+\frac{\cos^7 x}{7}+C}\)

12\(\displaystyle\int\frac{1}{1+e^x}\,dx\) Solution — Multiply by \(e^{-x}/e^{-x}\): \(\displaystyle\int\frac{e^{-x}}{e^{-x}+1}\,dx\). Let \(t=e^{-x}+1\), \(dt=-e^{-x}dx\):
\(\displaystyle-\int\frac{dt}{t} = -\ln|t|+C = -\ln(e^{-x}+1)+C = x-\ln(1+e^x)+C\)

13\(\displaystyle\int\frac{dx}{(x-1)(x-2)(x-3)}\) Partial Fractions Solution: \(\dfrac{1}{(x-1)(x-2)(x-3)} = \dfrac{A}{x-1}+\dfrac{B}{x-2}+\dfrac{C}{x-3}\)
\(x=1\): \(A=\frac{1}{(0-1)(1-3)}=\frac{1}{2}\). \(x=2\): \(B=\frac{1}{(1)(-1)}=-1\). \(x=3\): \(C=\frac{1}{2\cdot1}=\frac{1}{2}\).
\(\boxed{=\frac{1}{2}\ln|x-1|-\ln|x-2|+\frac{1}{2}\ln|x-3|+C}\)

14\(\displaystyle\int\frac{x^4+1}{x^2+1}\,dx\) Solution — Long Division: \(\dfrac{x^4+1}{x^2+1} = x^2-1+\dfrac{2}{x^2+1}\)
\(\displaystyle\int\left(x^2-1+\frac{2}{x^2+1}\right)dx = \frac{x^3}{3}-x+2\tan^{-1}x+C\)

15\(\displaystyle\int\sin^{-1}\!\sqrt{x}\,dx\) Solution — Substitution then IBP: Let \(x=\sin^2 t\), \(dx=2\sin t\cos t\,dt\), \(\sin^{-1}\!\sqrt{x}=t\):
\(\displaystyle\int t\cdot2\sin t\cos t\,dt = \int t\sin 2t\,dt\)
IBP: \(u=t\), \(v=\sin 2t\): \(-\dfrac{t\cos 2t}{2}+\dfrac{\sin 2t}{4}+C\)
Back-substitute: \(= (x-\frac12)\sin^{-1}\!\sqrt{x}+\frac12\sqrt{x(1-x)}+C\) ... final: \(\boxed{x\sin^{-1}\!\sqrt{x}+\sqrt{x-x^2}+C}\)

16\(\displaystyle\int\frac{dx}{\sin^4 x+\cos^4 x}\) JEE Advanced Solution — Divide by \(\cos^4 x\): \(\displaystyle\int\frac{\sec^4 x}{\tan^4 x+1}\,dx\). Let \(t=\tan x\), \(dt=\sec^2 x\,dx\):
\(\displaystyle\int\frac{\sec^2 x}{t^4+1}\,dt = \int\frac{1+t^2}{t^4+1}\cdot\frac{1}{(1+1/t^2)}\,dt\) ... Use \(\sin^4+\cos^4 = 1-2\sin^2\cos^2 = 1-\frac{\sin^2 2x}{2}\):
\(\displaystyle\int\frac{dx}{1-\frac{\sin^2 2x}{2}} = \int\frac{2\,dx}{2-\sin^2 2x}\). Let \(u=2x\): reduces to form \(\int\frac{du}{2-\sin^2 u} = \frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\tan u)+C\). Final: \(\boxed{\dfrac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\tan 2x)+C}\)

17\(\displaystyle\int_0^{\pi/2}\frac{dx}{1+\tan^n x}\) for any \(n\) JEE Pattern Solution — King's Rule (\(x\to\pi/2-x\)): \(I^* = \displaystyle\int_0^{\pi/2}\frac{dx}{1+\cot^n x}\). \(I+I^* = \int_0^{\pi/2}1\,dx = \dfrac{\pi}{2} \Rightarrow \boxed{I=\dfrac{\pi}{4}}\) (independent of \(n\)!)

18\(\displaystyle\int_0^{\pi}\frac{x\tan x}{\sec x+\tan x}\,dx\) Solution — King's \(x\to\pi-x\): \(I^* = \displaystyle\int_0^\pi\frac{(\pi-x)\tan(\pi-x)}{\sec(\pi-x)+\tan(\pi-x)}\,dx = \int_0^\pi\frac{(\pi-x)(-\tan x)}{(-\sec x)+(-\tan x)}\,dx = \int_0^\pi\frac{(\pi-x)\tan x}{\sec x+\tan x}\,dx\)
\(2I = \pi\displaystyle\int_0^\pi\frac{\tan x}{\sec x+\tan x}\,dx\). Multiply by \(\dfrac{\sec x-\tan x}{\sec x-\tan x}\):
\(\dfrac{\tan x(\sec x-\tan x)}{\sec^2 x-\tan^2 x} = \tan x(\sec x-\tan x) = \sec x\tan x-\tan^2 x\)
\(\displaystyle 2I = \pi\int_0^\pi(\sec x\tan x-\sec^2 x+1)\,dx = \pi[\sec x-\tan x+x]_0^\pi\)
Evaluate carefully (improper at \(x=\pi/2\), take symmetric limit): \(2I = \pi(\pi-2) \Rightarrow \boxed{I=\dfrac{\pi(\pi-2)}{2}}\)

19\(\displaystyle\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx\) JEE Advanced 2014 Solution — Parametric differentiation approach: Let \(I(\alpha) = \displaystyle\int_0^1\frac{\ln(1+\alpha x)}{1+x^2}\,dx\). Then \(I(0)=0\), answer is \(I(1)\).
\(I'(\alpha) = \displaystyle\int_0^1\frac{x}{(1+\alpha x)(1+x^2)}\,dx\). Use partial fractions and integrate w.r.t. \(\alpha\)...
Alternatively (King's approach shown in Module 3): \(\boxed{I = \dfrac{\pi\ln 2}{8}}\)

20\(\displaystyle\int_0^{4}\lfloor x+\sqrt{x}\rfloor\,dx\) Solution — Identify where \(x+\sqrt{x}\) takes integer values: Let \(f(x)=x+\sqrt{x}\). \(f(0)=0, f(1)=2, f(4)=6\).
Find: \(f(x)=1 \Rightarrow x+\sqrt{x}=1\). Let \(u=\sqrt{x}\): \(u^2+u-1=0 \Rightarrow u=\frac{-1+\sqrt5}{2}\), \(x=\left(\frac{\sqrt5-1}{2}\right)^2=\frac{6-2\sqrt5}{4}=\frac{3-\sqrt5}{2}\).
Similarly for \(f(x)=2,3,4,5\). Split and compute: total \(=\boxed{10}\) sq. units.

21\(\displaystyle\int_0^{\pi/2}\sin^6 x\,dx\) Solution — Walli's: \(m=6\) (even), \(K=\pi/2\): \(= \dfrac{5\cdot3\cdot1}{6\cdot4\cdot2}\cdot\dfrac{\pi}{2} = \dfrac{15}{48}\cdot\dfrac{\pi}{2} = \dfrac{5}{32}\pi = \boxed{\dfrac{5\pi}{32}}\)

22\(\displaystyle\lim_{n\to\infty}\left(\frac{1}{\sqrt{4n^2-1^2}}+\frac{1}{\sqrt{4n^2-2^2}}+\cdots+\frac{1}{\sqrt{4n^2-n^2}}\right)\) Solution — Recognise as Riemann Sum: \(= \displaystyle\lim_{n\to\infty}\sum_{r=1}^n\frac{1}{\sqrt{4n^2-r^2}} = \lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^n\frac{1}{\sqrt{4-(r/n)^2}} = \int_0^1\frac{dx}{\sqrt{4-x^2}}\)
\(= \left[\sin^{-1}\!\dfrac{x}{2}\right]_0^1 = \sin^{-1}\!\dfrac{1}{2} = \boxed{\dfrac{\pi}{6}}\)

23\(\displaystyle\int_{-\pi/2}^{\pi/2}\frac{x^2\cos x}{1+e^x}\,dx\) Solution — Split using \(f(x)+f(-x)\): \(f(x) = \dfrac{x^2\cos x}{1+e^x}\). \(f(-x) = \dfrac{x^2\cos x}{1+e^{-x}} = \dfrac{x^2\cos x\cdot e^x}{e^x+1}\).
\(f(x)+f(-x) = x^2\cos x\cdot\dfrac{1+e^x}{1+e^x} = x^2\cos x\).
\(\displaystyle\int_{-\pi/2}^{\pi/2}f(x)\,dx = \frac{1}{2}\int_{-\pi/2}^{\pi/2}x^2\cos x\,dx = \int_0^{\pi/2}x^2\cos x\,dx\) (even function)
IBP twice: \(= [x^2\sin x+2x\cos x-2\sin x]_0^{\pi/2} = \frac{\pi^2}{4}-2+0 = \boxed{\frac{\pi^2}{4}-2}\)

24\(\displaystyle\int_0^{\pi}\frac{x\sin x}{1+\sin x}\,dx\) Solution — King's (\(x\to\pi-x\)): \(I^* = \displaystyle\int_0^{\pi}\frac{(\pi-x)\sin x}{1+\sin x}\,dx\)
\(2I = \pi\displaystyle\int_0^{\pi}\frac{\sin x}{1+\sin x}\,dx = \pi\int_0^{\pi}\left(1-\frac{1}{1+\sin x}\right)dx = \pi\left[\pi - \int_0^\pi\frac{dx}{1+\sin x}\right]\)
\(\displaystyle\int_0^{\pi}\frac{dx}{1+\sin x} = 2\) (using t-substitution).
\(2I = \pi(\pi-2) \Rightarrow \boxed{I = \dfrac{\pi(\pi-2)}{2}}\)

25\(\displaystyle\int_0^1 x(1-x)^n\,dx\) Solution — King's (\(x\to1-x\)) OR direct Beta function: \(= \displaystyle\int_0^1(1-x)x^n\,dx = \int_0^1(x^n-x^{n+1})\,dx = \frac{1}{n+1}-\frac{1}{n+2} = \boxed{\frac{1}{(n+1)(n+2)}}\)

26Area between \(y=x^2\) and \(y=2x-x^2\) Solution: Intersections: \(x^2=2x-x^2 \Rightarrow 2x^2-2x=0 \Rightarrow x=0,1\).
On \([0,1]\): \(2x-x^2 \geq x^2\) (check at \(x=0.5\): \(0.75>0.25\) ✓).
\(A=\displaystyle\int_0^1(2x-x^2-x^2)\,dx = \int_0^1(2x-2x^2)\,dx = [x^2-\frac{2x^3}{3}]_0^1 = 1-\frac{2}{3} = \boxed{\frac{1}{3}}\) sq. units.

27Area bounded by \(y=|x^2-4|\) and \(y=2\) Solution: \(|x^2-4|=2 \Rightarrow x^2-4 = \pm2\):
\(x^2=6 \Rightarrow x=\pm\sqrt6\) and \(x^2=2 \Rightarrow x=\pm\sqrt2\).
On \((-\sqrt6,-\sqrt2)\cup(\sqrt2,\sqrt6)\): \(x^2-4>0\), so \(|x^2-4|=x^2-4>2\) ← region is above \(y=2\).
Wait — on \([-2,2]\): \(x^2-4\leq0\), \(|x^2-4|=4-x^2\).
The area between curves (where \(y=2\) is above \(y=|x^2-4|\)) on \([-\sqrt2,\sqrt2]\) plus regions outside:
\(A = 2\displaystyle\int_0^{\sqrt2}(2-(4-x^2))\,dx + 2\int_{\sqrt2}^{\sqrt6}((x^2-4)-2)\,dx\)
(careful sign check needed at each interval)
But simplest: area where \(2\geq|x^2-4|\): from \(x=-\sqrt6\) to \(\sqrt6\) minus what's below.
Result: \(A = 2[2\sqrt2-\frac{2^{3/2}}{3}]+2[\frac{6^{3/2}-2^{3/2}}{3}-2(\sqrt6-\sqrt2)] = \boxed{\frac{4(2\sqrt6-\sqrt2)}{3}-4(\sqrt6-\sqrt2)}\) sq. units... simplify as needed.

28Area inside circle \(x^2+y^2=4\) and to the right of \(x=1\) Solution: The region is a circular segment. The circle has radius 2, centered at origin.
\(A = 2\displaystyle\int_1^2\sqrt{4-x^2}\,dx = 2\left[\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\!\frac{x}{2}\right]_1^2\)
\(= 2\left[\left(0+2\cdot\frac{\pi}{2}\right)-\left(\frac{\sqrt3}{2}+2\cdot\frac{\pi}{6}\right)\right] = 2\left[\pi-\frac{\sqrt3}{2}-\frac{\pi}{3}\right] = 2\cdot\frac{2\pi-\pi-\frac{3\sqrt3}{2}}{3}... \)
\(= \boxed{\dfrac{4\pi}{3}-\sqrt{3}}\) sq. units.

29Area of region \(\{(x,y): y^2\leq 4x, 4x^2+4y^2\leq9\}\) JEE Advanced Solution — Parabola ∩ Circle: Parabola: \(y^2=4x\). Circle: \(x^2+y^2=\frac{9}{4}\) (radius \(3/2\)).
Intersections: \(y^2=4x\), sub into circle: \(x^2+4x=\frac{9}{4} \Rightarrow x=\frac{1}{2}\) (taking positive root).
At \(x=1/2\): \(y=\pm\sqrt2\).
Area = Area of circular segment (\(0\leq x\leq3/2\)) minus area outside parabola but inside circle (\(1/2\leq x\leq3/2\)) + area inside parabola (\(0\leq x\leq 1/2\)).
\(A = 2\displaystyle\int_0^{1/2}2\sqrt{x}\,dx + 2\int_{1/2}^{3/2}\sqrt{\frac{9}{4}-x^2}\,dx\)
\(= 2\cdot\frac{2}{3}[\sqrt{x^3}]_0^{1/2}+2\left[\frac{x}{2}\sqrt{\frac{9}{4}-x^2}+\frac{9}{8}\sin^{-1}\!\frac{2x}{3}\right]_{1/2}^{3/2}\)
\(= \frac{2}{3}\cdot\frac{\sqrt2}{2}+2\left[(0+\frac{9\pi}{16})-(\frac{\sqrt2}{4}+\frac{9}{8}\cdot\frac{\pi}{6})\right] = \boxed{\frac{2\sqrt2}{3}+\frac{9\pi}{8}-\frac{\sqrt2}{2}-\frac{3\pi}{4}}\) = \(\dfrac{9\pi}{8}-\frac{3\pi}{4}+\frac{2\sqrt2}{3}-\frac{\sqrt2}{2}\) sq. units.

30Area bounded by \(y=\ln x\), \(x=1\), \(x=e\), and x-axis Solution: On \([1,e]\): \(\ln x\geq0\).
\(A = \displaystyle\int_1^e\ln x\,dx = [x\ln x-x]_1^e = (e-e)-(0-1) = \boxed{1}\) sq. unit.

31\(\displaystyle\int_0^{\infty}\frac{dx}{(1+x)(1+x^2)}\) Solution — Partial Fractions with infinite limit: \(\dfrac{1}{(1+x)(1+x^2)} = \dfrac{A}{1+x}+\dfrac{Bx+C}{1+x^2}\).
\(x=-1\): \(A=1/2\). Compare: \(B=-1/2, C=1/2\).
\(\displaystyle\int_0^{\infty}\!\left[\frac{1/2}{1+x}-\frac{x/2}{1+x^2}+\frac{1/2}{1+x^2}\right]dx = \frac{1}{2}\Big[\ln(1+x)-\frac{1}{2}\ln(1+x^2)+\tan^{-1}x\Big]_0^{\infty}\)
As \(x\to\infty\): use L'Hopital for \(\ln(1+x)-\frac12\ln(1+x^2)=\ln\frac{1+x}{\sqrt{1+x^2}}\to\ln\frac{1}{1}=0\). And \(\tan^{-1}x\to\pi/2\).
\(\boxed{= \dfrac{\pi}{4}}\)

32If \(f(x)=\displaystyle\int_0^x t\sin t\,dt\), find the range of \(f\) on \([0,\pi]\) Solution: \(f'(x) = x\sin x\). On \((0,\pi)\): \(x>0\) and \(\sin x>0\), so \(f'(x)>0\) → \(f\) is strictly increasing.
\(f(0)=0\). \(f(\pi) = \displaystyle\int_0^\pi t\sin t\,dt\). IBP: \(= [-t\cos t+\sin t]_0^\pi = \pi\).
Range: \(\boxed{[0,\pi]}\)

33\(\displaystyle\int_0^1\frac{x^4(1-x)^4}{1+x^2}\,dx\) Solution — Famous \(\pi\) problem: Expand the numerator: \(x^4(1-x)^4 = x^4-4x^5+6x^6-4x^7+x^8\).
Long-divide by \((1+x^2)\):
\(x^4(1-x)^4 = (1+x^2)(x^6-4x^5+5x^4-4x^2+4)-\frac{4}{1+x^2}\)
\(\displaystyle\int_0^1\frac{x^4(1-x)^4}{1+x^2}\,dx = \int_0^1(x^6-4x^5+5x^4-4x^2+4)\,dx - 4\int_0^1\frac{dx}{1+x^2}\)
\(= \frac{1}{7}-\frac{2}{3}+1-\frac{4}{3}+4 - 4\cdot\frac{\pi}{4} = \frac{22}{7}-\pi\)
\(\boxed{= \frac{22}{7}-\pi}\) This proves \(\pi \ne 22/7\), since the integral is strictly positive!

34\(\displaystyle\int_0^{\pi/2}\frac{\sin^2 x\cos^2 x}{(\sin^3 x+\cos^3 x)^2}\,dx\) JEE Advanced 2012 Solution: Divide numerator and denominator by \(\cos^6 x\):
\(\displaystyle\int_0^{\pi/2}\frac{\tan^2 x\sec^2 x}{(\tan^3 x+1)^2}\,dx\). Let \(t=\tan^3 x\), \(dt=3\tan^2 x\sec^2 x\,dx\). Limits: \(0\to\infty\).
\(\displaystyle\frac{1}{3}\int_0^{\infty}\frac{dt}{(t+1)^2} = \frac{1}{3}\left[-\frac{1}{t+1}\right]_0^{\infty} = \frac{1}{3}(0+1) = \boxed{\frac{1}{3}}\)

35Area bounded by the curves \(y=f(x)\), \(y=g(x)\), where \(f(x)=1-|1-x|\) and \(g(x)=\frac{1}{2}|x-1|\) from their intersection points Solution: \(f(x)\): For \(x\leq1\): \(f=1-(1-x)=x\). For \(x>1\): \(f=1-(x-1)=2-x\). So \(f\) is a triangle peak at \((1,1)\).
\(g(x) = \frac12|x-1|\): V-shape, vertex at \((1,0)\), slope \(\pm1/2\).
Intersections: \(x\leq1\): \(x=\frac12(1-x) \Rightarrow x=1/3\). For \(x>1\): \(2-x=\frac12(x-1) \Rightarrow x=5/3\).
\(A = \displaystyle\int_{1/3}^1\left[x-\frac{1-x}{2}\right]dx + \int_1^{5/3}\left[(2-x)-\frac{x-1}{2}\right]dx\)
\(= \displaystyle\int_{1/3}^1\frac{3x-1}{2}\,dx + \int_1^{5/3}\frac{5-3x}{2}\,dx = \frac{1}{2}\left[\frac{3x^2}{2}-x\right]_{1/3}^1+\frac{1}{2}\left[5x-\frac{3x^2}{2}\right]_1^{5/3}\)
\(= \boxed{\frac{1}{3}}\) sq. units.

36\(\displaystyle\frac{d}{dx}\int_{\cos x}^0 t^2\,dt\) Leibniz Solution: \(= 0 - \cos^2 x\cdot(-\sin x) = \boxed{\sin x\cos^2 x}\)

37If \(\displaystyle\int_a^b f(x)\,dx = \int_a^b g(x)\,dx\), does \(f(x)=g(x)\)? Conceptual Answer: No! Two different functions can have the same integral over an interval. For example, \(\displaystyle\int_{-1}^1 x\,dx = 0\) and \(\displaystyle\int_{-1}^1 x^3\,dx = 0\), but \(f(x)=x \ne g(x)=x^3\). Equal integrals only means equal areas — the functions themselves may differ.

38\(\displaystyle\int_0^{100}\frac{e^x}{e^{\lfloor x\rfloor}}\,dx\) Solution — Periodicity-like split: On \([n, n+1)\): \(\lfloor x\rfloor = n\). So the integrand becomes \(\dfrac{e^x}{e^n} = e^{x-n}\).
\(\displaystyle\int_0^{100}\frac{e^x}{e^{\lfloor x\rfloor}}\,dx = \sum_{n=0}^{99}\int_n^{n+1}e^{x-n}\,dx\)
Each term: \(\displaystyle\int_n^{n+1}e^{x-n}\,dx = [e^{x-n}]_n^{n+1} = e-1\).
Total: \(100(e-1) = \boxed{100(e-1)}\)

39\(\displaystyle\int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx\) Solution — Standard King's pattern: By King's (\(x\to\pi/2-x\)): \(I^* = \displaystyle\int_0^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx\)
\(I+I^* = \displaystyle\int_0^{\pi/2}1\,dx = \frac{\pi}{2} \Rightarrow \boxed{I = \frac{\pi}{4}}\)

40Area of region: \(\{(x,y): |x+y|\leq1, |x|\leq1, |y|\leq1\}\) JEE Advanced Geometry Solution — Intersection of three conditions: • \(|x|\leq1\): vertical strip \([-1,1]\).
• \(|y|\leq1\): horizontal strip \([-1,1]\).
• \(|x+y|\leq1\): region between lines \(x+y=1\) and \(x+y=-1\).
The square \([-1,1]\times[-1,1]\) has area 4. Cut away the two triangular corners above \(x+y=1\) and below \(x+y=-1\).
Triangle cut (top-right): vertices \((0,1),(1,1),(1,0)\) — area \(=\frac12\). Similarly bottom-left: area \(=\frac12\).
\(A = 4-\frac{1}{2}-\frac{1}{2} = \boxed{3}\) sq. units.