Solve the Following Question.(2 Marks)

Question 12 Marks

$\int_2^3 \frac{x}{x^2-1} d x$

Answer

Let $I =\int_2^3 \frac{x}{x^2-1} \cdot d x$
Put $x ^2-1= t$
$
\begin{aligned}
& \therefore 2 x \cdot dx = dt \\
& \therefore x \cdot dx =\frac{1}{2} \cdot d t
\end{aligned}
$
When $x=2, t=2^2-1=3$
When $x=3, t=3^2-1=8$
$
\begin{aligned}
& \therefore I =\int_3^8 \frac{1}{ t } \cdot \frac{ dt }{2} \\
& =\frac{1}{2} \int_3^8 \frac{ dt }{ t } \\
& =\frac{1}{2}[\log | t |]_3^8 \\
& =\frac{1}{2}(\log 8-\log 3) \\
& \therefore I =\frac{1}{2} \log \left(\frac{8}{3}\right) .
\end{aligned}
$

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Question 22 Marks

$\int_1^3 x^2 \log x d x$

Answer

$
\begin{aligned}
& \int_1^3 x^2 \log x d x=\int_1^3(\log x) \cdot x^2 d x \\
= & {\left[(\log x) \int x^2 d x\right]_1^3-\int_1^3\left[\frac{d}{d x}(\log x) \int x^2 d x\right] d x } \\
= & {\left[(\log x)\left(\frac{x^3}{3}\right)\right]_1^3-\int_1^3 \frac{1}{x} \times \frac{x^3}{3} d x } \\
= & \frac{1}{3}\left[x^3 \log x\right]_1^3-\frac{1}{3} \int_1^3 x^2 d x \\
= & \frac{1}{3}[27 \log 3-0]-\frac{1}{3}\left[\frac{x^3}{3}\right]_1^3 \quad \ldots[\because \log 1=0] \\
= & 9 \log 3-\frac{1}{9}(27-1) \\
= & 9 \log 3-\frac{26}{9} .
\end{aligned}
$

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Question 32 Marks

$\int_0^9 \frac{1}{1+\sqrt{x}} d x$

Answer

Let $I=\int_0^9 \frac{1}{1+\sqrt{x}} d x$
Put $\sqrt{x}=t$, i.e. $x=t^2$
$
\begin{aligned}
& \therefore d x=2 t d t \\
& \text { When } x=0, t=0 \\
& \text { When } x=9, t=3
\end{aligned}
$
$
\therefore{ }^3 1
$
$
\begin{aligned}
& =2 \int_0^3\left[\frac{(1+t)-1}{1+t}\right] d t \\
& =2 \int_0^3\left(1-\frac{1}{1+t}\right) d t \\
& =2[t-\log |1+t|]_0^3 \\
& =2[(3-\log 4)-0] \\
& =6-2 \log 4=6-4 \log 2 .
\end{aligned}
$

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Question 42 Marks

$\int_0^4 \frac{1}{\sqrt{x^2+2 x+3}} d x$

Answer

$
\int_0^4 \frac{1}{\sqrt{x^2+2 x+3}} d x
$
$
\int_0^4 \frac{1}{\sqrt{\left(x^2+2 x+1\right)+2}} d x
$
$
\begin{aligned}
& \int_0^4 \frac{1}{\sqrt{(x+1)^2+(\sqrt{2})^2}} \\
& =\left[\log \left|(x+1)+\sqrt{\left.(x+1)^2+\sqrt{2}\right)^2}\right|\right]_0^4 \\
& =\left[\log \left|(x+1)+\sqrt{x^2+2 x+3}\right|\right]_0^4 \\
& =\log (5+\sqrt{27})-\log (1+\sqrt{3}) \\
& =\log \left(\frac{5+3 \sqrt{3}}{1+\sqrt{3}}\right)
\end{aligned}
$

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Question 52 Marks

$\int_2^3 \frac{x}{x^2+1} d x$

Answer

Let $I=\int_2^3 \frac{x}{x^2+1} d x$
Put $x^2+1=t \quad \therefore 2 x d x=d t$
$
\therefore x d x=\frac{d t}{2}
$
When $x=2, t=4+1=5$
When $x=3, t=9+1=10$
$
\begin{aligned}
\therefore I & =\int_5^{10} \frac{1}{t} \cdot \frac{d t}{2}=\frac{1}{2} \int_5^{10} \frac{1}{t} d t \\
& =\frac{1}{2}[\log |t|]_5^{10}
\end{aligned}
$
$
\begin{aligned}
& =\frac{1}{2}(\log 10-\log 5)=\frac{1}{2} \log \left(\frac{10}{5}\right) \\
& =\frac{1}{2} \log 2=\log \sqrt{2} .
\end{aligned}
$

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Question 62 Marks

Evaluate the following integrals : $\int_0^1 x(1-x)^5 d x$

Answer

We use the property
$
\begin{aligned}
\int_0^a f(x) d x=\int_0^a f(a-x) d x \\
\therefore \int_0^1 x(1-x)^5 d x & =\int_0^1(1-x)(1-1+x)^5 d x \\
=\int_0^1(1-x) x^5 d x=\int_0^1\left(x^5-x^6\right) d x \\
=\int_0^1 x^5 d x-\int_0^1 x^6 d x \\
=\left[\frac{x^6}{6}\right]_0^1-\left[\frac{x^7}{7}\right]_0^1 \\
=\frac{1}{6}(1-0)-\frac{1}{7}(1-0) \\
=\frac{1}{6}-\frac{1}{7}=\frac{1}{42} .
\end{aligned}
$

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Question 72 Marks

Evaluate the following integrals : $\int_0^1 \log \left(\frac{1}{x}-1\right) d x$

Answer

Let $I=\int_0^1 \log \left(\frac{1}{x}-1\right) d x=\int_0^1 \log \left(\frac{1-x}{x}\right) d x$
We use the property, $\int_0^a f(x) d x=\int_0^a f(a-x) d x$
$
\begin{aligned}
\therefore 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 \\
& =\int_0^1-\log \left(\frac{1-x}{x}\right) d x=-\int_0^1 \log \left(\frac{1-x}{x}\right) d x
\end{aligned}
$
$
\begin{aligned}
& \therefore I=-I \\
& \therefore 2 I=0 \quad \therefore I=0
\end{aligned}
$
Hence, $\int_0^1 \log \left(\frac{1}{x}-1\right) d x=0$.

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Question 82 Marks

Evaluate the following integrals : $\int_{-9}^9 \frac{x^3}{4-x^2} d x$

Answer

Let $I=\int_{-9}^9 \frac{x^3}{4-x^2} d x$
Let $f(x)=\frac{x^3}{4-x^2}$
$
\therefore f(-x)=\frac{(-x)^3}{4-(-x)^2}=\frac{-x^3}{4+x^2}=-f(x)
$
$\therefore f$ is an odd function.
$\therefore \int_{-9}^9 f(x) d x=0$
i.e. $\int_{-9}^9 \frac{x^3}{4-x^2} d x=0$.

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Question 92 Marks

Evaluate the following definite integrals : If $\int_0^a(2 x+1) d x=2$, find the real values of ' $a$ '.

Answer

$
\begin{aligned}
& \text { Let } I =\int_0^a(2 x+1) d x \\
& =\left[2 \cdot \frac{x^2}{2}+x\right]_0^a \\
& = a ^2+ a -0 \\
& = a ^2+ a \\
& \therefore I =2 \text { gives } a ^2+ a =2 \\
& \therefore a ^2+ a -2=0 \\
& \therefore( a +2)( a -1)=01 \\
& \therefore a +2=0 \text { or } a -1=0 \\
& \therefore a =-2 \text { or } a =1 .
\end{aligned}
$

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Question 102 Marks

Evaluate the following definite integrals : $\int_1^2 \frac{d x}{x^2+6 x+5}$

Answer

$
\begin{aligned}
& =\int_1^2 \frac{d x}{\left(x^2+6 x+9\right)-4} \frac{d x}{x^2+6 x+5} \\
& =\int_1^2 \frac{1}{(x+3)^2-(2)^2} d x \\
& =\frac{1}{2(2)}\left[\log \left|\frac{x+3-2}{x+3+2}\right|\right]_1^2=\frac{1}{4}\left[\log \left|\frac{x+1}{x+5}\right|\right]_1^2 \\
& =\frac{1}{4}\left[\log \frac{3}{7}-\log \frac{2}{6}\right] \\
& =\frac{1}{4} \log \left(\frac{3}{7} \times \frac{6}{2}\right)=\frac{1}{4} \log \left(\frac{9}{7}\right) .
\end{aligned}
$

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Question 112 Marks

Evaluate the following definite integrals : $\int_0^1 \frac{x^2+3 x+2}{\sqrt{x}} d x$

Answer

$
\begin{aligned}
& \int_0^1 \frac{x^2+3 x+2}{\sqrt{x}} d x \\
= & \int_0^1\left(\frac{x^2}{\sqrt{x}}+\frac{3 x}{\sqrt{x}}+\frac{2}{\sqrt{x}}\right) d x \\
= & \int_0^1\left(x^{\frac{3}{2}}+3 x^{\frac{1}{2}}+2 x^{-\frac{1}{2}}\right) d x \\
= & {\left[\frac{x^{\frac{5}{2}}}{5 / 2}+3\left(\frac{x^{\frac{3}{2}}}{3 / 2}\right)+2\left(\frac{x^{\frac{1}{2}}}{1 / 2}\right)\right]_0^1 } \\
= & {\left[\frac{2}{5} x^{\frac{5}{2}}+2 x^{\frac{3}{2}}+4 x^{\frac{1}{2}}\right]_0^1 } \\
= & {\left[\frac{2}{5}(1)^{\frac{5}{2}}+2(1)^{\frac{3}{2}}+4(1)^{\frac{1}{2}}\right]-(0+0+0) } \\
= & \frac{2}{5}+2+4=\frac{32}{5} .
\end{aligned}
$

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Question 122 Marks

Evaluate the following definite integrals : $\int_2^3 \frac{x}{x^2-1} d x$

Answer

$
\begin{aligned}
& \int_2^3 \frac{x}{x^2-1} d x \\
& =\frac{1}{2} \int_2^3 \frac{2 x}{x^2-1} d x \\
& =\frac{1}{2}\left[\log \left|x^2-1\right|\right]_2^3 \quad \ldots \quad\left[\because \frac{d}{d x}\left(x^2-1\right)=2 x\right. \text { and } \\
& \left.\quad \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right] \\
& =\frac{1}{2}[\log (9-1)-\log (4-1)]=\frac{1}{2} \log \left(\frac{8}{3}\right) .
\end{aligned}
$

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Question 132 Marks

Evaluate the following definite integrals : $\int_1^3 \log x d x$

Answer

$
\begin{aligned}
& \int_1^3 \log x d x=\int_1^3(\log x) \cdot 1 d x \\
= & {\left.[(\log x)] \int 1 d x\right]_1^3-\int_1^3\left[\frac{d}{d x}(\log x) \int 1 d x\right] d x } \\
= & {[(\log x) x]_1^3-\int_1^3 \frac{1}{x} \times x d x } \\
= & (3 \log 3-\log 1)-\int_1^3 1 d x \\
= & 3 \log 3-[x]_1^3 \\
= & \log 3^3-(3-1) \\
= & \log 27-2 .
\end{aligned}
$

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Question 142 Marks

Evaluate the following definite integrals : $\int_1^2 \frac{3 x}{9 x^2-1} d x$

Answer

Let $I =\int_1^2 \frac{3 x}{9 x^2-1} d x=\int_1^2 \frac{3 x}{(3 x)^2-1} d x$
Put $3 x = t$
$\therefore 3 dx = dt$
$\therefore dx =\frac{d t}{3}$
When $x=1, t=3 \times 1=3$
When $x=2, t=3 \times 2=6$
$
\begin{aligned}
\therefore I & =\int_3^6 \frac{t}{t^2-1} \cdot \frac{d t}{3}=\frac{1}{6} \int_3^6 \frac{2 t}{t^2-1} d t \\
& =\frac{1}{6}\left[\log \left|t^2-1\right|\right]_3^6 \quad \ldots\left[\because \frac{d}{d t}\left(t^2-1\right)=2 t\right] \\
& =\frac{1}{6}[\log 35-\log 8] \\
& =\frac{1}{6} \log \left(\frac{35}{8}\right) .
\end{aligned}
$
Alternative Method :
$
\begin{aligned}
& \int_1^2 \frac{3 x}{9 x^2-1} d x \\
& =\frac{1}{6} \int_1^2 \frac{18 x}{9 x^2-1} d x \\
& =\frac{1}{6}\left[\log \left|9 x^2-1\right|\right]_1^2 \\
& =\frac{1}{6}[\log 35-\log 8] \\
& =\frac{1}{6} \log \left(\frac{35}{8}\right) .
\end{aligned}
$

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Maths (commerce) Solve the Following Question.(2 Marks)

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