Let \( \mathrm{I} =\int_{0}^{1} \frac{2 x}{1+x^{2}} \mathrm{~d} x \)
\( =\left[\log \left(1+x^{2}\right)\right]_{0}^{1}=\log (2)-\log (1) \)
\(\mathbf{I} =\log 2 \)

\( \mathrm{f}(x)=|x-1|=\left\{\begin{array}{cc} -(x-1) \text { if } x<1 \\ x-1 & \text { if } x \geq 1 \end{array}\right. \)
\( I=\int_{0}^{1}|x-1| d x \)
\( =\int_{0}^{1}-(x-1) d x=-\left(\frac{x^{2}}{2}-x\right)_{0}^{1} \)
\( =-\left(\frac{1}{2}-1-0\right)=-\left(\frac{-1}{2}\right)=\frac{1}{2} \)

\( I =\int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x \)
\( =\int_{0}^{1} \log \left(\frac{1-x}{x}\right) d x\) ..........(1)
\( \int_{0}^{a} f(x) d x =\int_{0}^{a} f(\mathrm{a}-x) d x \)
\(\therefore \mathrm{I} =\int_{0}^{1} \log \left(\frac{1-1+x}{1-x}\right) d x \)
\( =\int_{0}^{1} \log \left(\frac{x}{1-x}\right) d x \) ..............(2)
Add (1) & (2),
\( 2 \mathrm{I} =\int_{0}^{1}\left[\log \left(\frac{1-x}{x}\right)+\log \left(\frac{x}{1-x}\right)\right] d x \)
\(=\int_{0}^{1} \log \left(\frac{1-x}{x} \cdot \frac{x}{1-x}\right) d x=\int_{0}^{1} \log 1 \mathrm{dx} \)
2I = 0
I = 0
\(\therefore \int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x=0 \)

Since the solid is generated by revolving about the y axis, rewrite \(\mathrm{y}=\sqrt{x}\)
as x = y²
Volume is given by \( \mathbf{V} =\int_{c}^{d} \pi x^{2} d y \)
\( =\int_{0}^{2} \pi y^{4} d y=\left[\frac{\pi y^{5}}{5}\right]_{0}^{2} \)
\(V=\frac{32 \pi}{5}\)cube units.

For all x, \(2 \leq x \leq 3\) the curve lieg below the x axis 

Required area \(=\left|\int_{0}^{3} y d x\right|\)
\( =\int_{2}^{3}\left(x^{2}-5 x+4\right) d x \)
\( =\left[\frac{x^{3}}{3}-\frac{5 x^{2}}{2}+4 x\right]_{2}^{3} \)
\( =\left|\left(9-\frac{45}{2}+12\right)-\left(\frac{8}{3}-\frac{20}{2}+8\right)\right|\)
\(=\left|-\frac{13}{6}\right|=\frac{13}{6}\)
Required Area = \(\frac{13}{6}\)sq. units

We know that if a, b and c are in A.P
then \(\mathrm{b}=\frac{a+c}{2}\)
\(f(x)=\frac{\sin x+\cos x}{2}\)
Then \( \int f(x) d x =\int \frac{\sin x+\cos x}{2} d x \)
\( =\frac{1}{2}[-\cos x+\sin x]+c \)

\( I=\int_{0}^{\pi / 2} \frac{\tan ^{7} x}{\cot ^{7} x+\tan ^{7} x} d x \) .............. (1)
\( I=\int_{0}^{\pi / 2} \frac{\tan ^{7}(\pi / 2-x)}{\cot ^{7}(\pi / 2-x)+\tan ^{7}(\pi / 2-x)} d x \)
\( I=\int_{0}^{\pi / 2} \frac{\cot ^{7} x}{\tan ^{7} x+\cot ^{7} x} d x \) .............. (2)
Add (1) & (2),
\( 2 \mathrm{I}=\int_{0}^{\pi / 2} \frac{\tan ^{7} x+\cot ^{7} x}{\tan ^{7} x+\cot ^{7} x} d x \)
\( 2 \mathrm{I}=\int_{0}^{\pi / 2} d x=[x]_{0}^{\pi / 2} \)
\( 2 \mathrm{I}=\frac{\pi}{2} \)
\( \mathrm{I}=\frac{\pi}{4}\)

\( \mathrm{I} =\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x \) ............. (1)
\(\int_{0}^{a} f(x) d x =\int_{0}^{a} f(a-x) d x \)
\(\mathrm{I} =\int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{a+a-x}} d x \)
\(\mathrm{I} =\int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x \) ............. (2)
(1)+(2) \( =\int_{0}^{a} \frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x \)
\( =\int_{0}^{a} d x=[x]_{0}^{a} \)
\(2 \mathrm{I} =\mathrm{a} \)
\(\mathrm{I} =\frac{a}{2}\)

\( I =\int_{0}^{2 \pi} \frac{1}{1+e^{\sin x}} d x \)
\(\int_{0}^{20} f(x) d x =\int_{0}^{a}[f(x)+f(2 \mathrm{a}-x)] d x \)
\( =\int_{0}^{\pi}\left[\frac{1}{1+e^{\sin x}}+\frac{1}{1+e^{-\sin x}}\right] d x \)
\( =\int_{0}^{\pi}\left[\frac{1}{1+e^{\sin x}}+\frac{e^{\sin x}}{1+e^{\sin x}}\right] d x \)
\( =\int_{0}^{\pi} \frac{1+e^{\sin x}}{1+e^{\sin x}} d x=[x]_{0}^{\pi} \)
\(\int_{0}^{2 \pi} \frac{1}{1+e^{\sin x}} d x =\pi \)

\( \mathrm{I} =\int_{1}^{3} \frac{1}{x(1+\log x)} d x \)
\(\mathrm{t} =1+\log x \)
\(\mathrm{dt} =\frac{1}{x} \mathrm{dx} \)
\(\mathrm{I} =\int_{1}^{1+\log 3} \frac{d t}{t} \)
\( =[\log t]_{1}^{1+\log 3} \)
\( =\log (1+\log 3)-\log 1 \)
\( =\log (1+\log 3) \) 

\( I =\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}} \times \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}} \)
\( =\int_{0}^{1} \frac{(1+x)^{1 / 2}+x^{1 / 2}}{1+x-x} d x \)
\( =\left[\frac{(1+x)^{3 / 2}}{\frac{3}{2}+\frac{x^{3 / 2}}{3}}\right]_{0}^{1}=\frac{2}{3}[2 \sqrt{2}+1-1] \)
\( I=\frac{4 \sqrt{3}}{3} \)

Let I = \( \int_{0}^{\pi / 2} x \cos x d x\)
Applying the rule of integration by parts we get
\( I =[x \sin x-\sin x]_{0}^{\pi / 2} \)
\( =\frac{\pi}{2} \sin \frac{\pi}{2}-\sin \frac{\pi}{2} \)
\(I =\frac{\pi}{2}-1 \)

\(\frac { \pi }{ 4 } \)

\(\frac { 28 }{ 3 } \)

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