Solve the Following Question.(3 Marks)

Question 13 Marks

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

Answer

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

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Question 23 Marks

$\int_1^2 e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^2}\right) d x$

Answer

Let $I=\int_1^2 e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^2}\right) d x$
Put $2 x=t \quad \therefore 2 d x=d t$
$\therefore d x=\frac{d t}{2}$ and $x=\frac{t}{2}$
When $x=1, t=2$
When $x=2, t=4$
$
\therefore I=\int_2^4 e^t\left(\frac{2}{t}-\frac{2}{t^2}\right) \frac{d t}{2}=\frac{1}{2} \int_2^4 e^t\left(\frac{2}{t}-\frac{2}{t^2}\right) d t
$
Let $f(t)=\frac{2}{t}$
Then $f^{\prime}(t)=2\left(-\frac{1}{t^2}\right)=\frac{-2}{t^2}$
$
\begin{aligned}
\therefore I & =\frac{1}{2} \int_2^4 e^t\left[f(t)+f^{\prime}(t)\right] d t \\
& =\frac{1}{2}\left[e^t \cdot f(t)\right]_2^4=\frac{1}{2}\left[e^t \cdot \frac{2}{t}\right]_2^4 \\
& =\frac{1}{2}\left[e^4 \times \frac{2}{4}-e^2 \times \frac{2}{2}\right] \\
& =\frac{e^4}{4}-\frac{e^2}{2} .
\end{aligned}
$

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Question 33 Marks

$\int_0^1 e^{x^2} \cdot x^3 d x$

Answer

Let $I=\int_0^1 e^{x^2} \cdot x^3 d x=\int_0^1 e^{x^2} \cdot x^2 \cdot x d x$
Put $x^2=t \quad \therefore 2 x d x=d t$
$\therefore x d x=\frac{d t}{2}$
When $x=0, t=0$
When $x=1, t=1$
$
\begin{aligned}
\therefore I & =\int_0^1 e^t \cdot t \cdot \frac{d t}{2}=\frac{1}{2} \int_0^1 t e^t d t \\
& =\frac{1}{2}\left\{\left[t \int e^t d t\right]_0^1-\int_0^1\left[\frac{d}{d t}(t) \int e^t d t\right] d t\right\} \\
& =\frac{1}{2}\left[t e^t\right]_0^1-\frac{1}{2} \int_0^1 1 \cdot e^t d t \\
& =\frac{1}{2}(e-0)-\frac{1}{2}\left[e^t\right]_0^1 \\
& =\frac{e}{2}-\frac{1}{2}(e-1) \\
& =\frac{e}{2}-\frac{e}{2}+\frac{1}{2}=\frac{1}{2}
\end{aligned}
$

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Question 43 Marks

$\int_1^2 \frac{d x}{x(1+\log x)^2}$

Answer

$
\text { Let } \begin{aligned}
I & =\int_1^2 \frac{d x}{x(1+\log x)^2} \\
& =\int_1^2 \frac{1}{(1+\log x)^2} \cdot \frac{1}{x} d x
\end{aligned}
$
Put $1+\log x=t \quad \therefore \frac{1}{x} d x=d t$
When $x=1, t=1+\log 1=1+0=1$
When $x=2, t=1+\log 2$
$
\begin{aligned}
\therefore I & =\int_1^{1+\log 2} \frac{1}{t^2} d t=\int_1^{1+\log 2} t^{-2} d t \\
& =\left[\frac{t^{-1}}{-1}\right]_1^{1+\log 2}=-\left[\frac{1}{t}\right]_1^{1+\log 2} \\
& =-\left[\frac{1}{1+\log 2}-1\right] \\
& =-\left[\frac{1-(1+\log 2)}{1+\log 2}\right]=\frac{\log 2}{1+\log 2} .
\end{aligned}
$

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Question 53 Marks

$\int_2^4 \frac{x}{x^2+1} d x$

Answer

$
\text { Let } I =\int_2^4 \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{ dt }{2}
\end{aligned}
$
When $x=2, t=2^2+1=5$
When $x=4, t=42+1=17$
$
\begin{aligned}
& \therefore I =\int_5^{17} \frac{1}{ t } \cdot \frac{ dt }{2} \\
& =\frac{1}{2} \int_5^{17} \frac{ dt }{ t } \\
& =\frac{1}{2}[\log | t |]_5^{17} \\
& =\frac{1}{2}(\log 17-\log 5) \\
& \therefore I =\frac{1}{2} \log \left(\frac{17}{5}\right) .
\end{aligned}
$

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Question 63 Marks

$\int_3^5 \frac{d x}{\sqrt{x+4}+\sqrt{x-2}}$

Answer

$
\begin{aligned}
& \int_3^5 \frac{d x}{\sqrt{x+4}+\sqrt{x-2}} \\
& =\int_3^5 \frac{1}{\sqrt{x+4}+\sqrt{x-2}} \times \frac{\sqrt{x+4}-\sqrt{x-2}}{\sqrt{x+4}-\sqrt{x-2}} d x \\
& =\int_3^5 \frac{\sqrt{x+4}-\sqrt{x-2}}{x+4-x+2} d x \\
& =\frac{1}{6} \int_3^5\left[(x+4)^{\frac{1}{2}}-(x-2)^{\frac{1}{2}}\right] d x \\
& =\frac{1}{6}\left[\frac{(x+4)^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}-\frac{(x-2)^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]_3^5 \\
& =\frac{1}{9}\left[(x+4)^{\frac{3}{2}}-(x-2)^{\frac{3}{2}}\right]_3^5 \\
& =\frac{1}{9}\left[\left(9^{\frac{3}{2}}-3^{\frac{3}{2}}\right)-\left(7^{\frac{3}{2}}-1\right)\right] \\
& =\frac{1}{9}(27-3 \sqrt{3}-7 \sqrt{7}+1) \\
& =\frac{1}{9}(28-3 \sqrt{3}-7 \sqrt{7}) .
\end{aligned}
$

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Question 73 Marks

Evaluate the following integrals : $\int_1^2 \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x$

Answer

$
\begin{aligned}
& \text { Let } I =\int_1^2 \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} \cdot d x \\
& =\int_1^2 \frac{\sqrt{1+2-x}}{\sqrt{3-(1+2-x)}+\sqrt{1+2-x}} \cdot d x \quad \ldots\left[\because \int_{ a }^{ b } f(x) \cdot d x=\int_{ a }^{ b } f( a + b -x) \cdot d x\right] \\
& \therefore I =\int_1^2 \frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} \cdot d x
\end{aligned}
$
Adding (i) and (ii), we get
$
\begin{aligned}
& 2 I =\int_1^2 \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} \cdot d x+\int_1^2 \frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} \cdot d x \\
& =\int_1^2 \frac{\sqrt{x}+\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} \cdot d x \\
& =\int_1^2 1 \cdot d x \\
& =[x]_1^2 \\
& \therefore 2 I =2-1=1 \\
& \therefore I =\frac{1}{2} .
\end{aligned}
$

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Question 83 Marks

Evaluate the following integrals : $\int_2^5 \frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}} d x$

Answer

Let $I=\int_2^5 \frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}} d x$
We use the property, $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$.
Hence in $I$, we change $x$ by $2+5-x$.
$
\begin{aligned}
\therefore I & =\int_2^5 \frac{\sqrt{2+5-x}}{\sqrt{2+5-x}+\sqrt{7-2-5+x}} d x \\
& =\int_2^5 \frac{\sqrt{7-x}}{\sqrt{7-x}+\sqrt{x}} d x
\end{aligned}
$
Adding (1) and (2), we get
$
\begin{array}{rl}
2 I & =\int_2^5 \frac{\sqrt{7-x}}{\sqrt{x}+\sqrt{7-x}} d x+\int_2^5 \frac{\sqrt{x}}{\sqrt{7-x}+\sqrt{x}} d x \\
& =\int_2^5 \frac{\sqrt{x}+\sqrt{7-x}}{\sqrt{x}+\sqrt{7-x}} d x \\
& =\int_2^5 1 d x=[x]_2^5=5-2=3 \\
\therefore & I=\frac{3}{2}
\end{array}
$
Hence, $\int_2^5 \frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}} d x=\frac{3}{2}$.

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Question 93 Marks

Evaluate the following integrals : $\int_1^3 \frac{\sqrt[3]{x+5}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}} d x$

Answer

Let $I=\int_1^3 \frac{\sqrt[3]{x+5}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}} d x$
We use the property, $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$
Hence in $I$, we replace $x$ by $1+3-x$.
$
\begin{aligned}
\therefore I & =\int_1^3 \frac{\sqrt[3]{1+3-x+5}}{\sqrt[3]{1+3-x+5}+\sqrt[3]{9-1-3+x}} d x \\
& =\int_1^3 \frac{\sqrt[3]{9-x}}{\sqrt[3]{9-x}+\sqrt[3]{x+5}} d x
\end{aligned}
$
Adding (1) and (2), we get
$
\begin{aligned}
2 I & =\int_1^3 \frac{\sqrt[3]{x+5}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}} d x+\int_1^3 \frac{\sqrt[3]{9-x}}{\sqrt[3]{9-x}+\sqrt[3]{x+5}} d x \\
& =\int_1^3 \frac{\sqrt[3]{x+5}+\sqrt[3]{9-x}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}} d x \\
& =\int_1^3 1 d x=[x]_1^3 \\
& =3-1=2 \\
\therefore & I=1
\end{aligned}
$
Hence, $\int_1^3 \frac{\sqrt[3]{x+5}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}} d x=1$.

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Question 103 Marks

Evaluate the following integrals : $\int_0^a x^2(a-x)^{3 / 2} d x$

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^a x^2(a-x)^{\frac{3}{2}} d x=\int_0^a(a-x)^2(a-a+x)^{\frac{3}{2}} d x \\
& =\int_0^a\left(a^2-2 a x+x^2\right) x^{\frac{3}{2}} d x \\
& =\int_0^a\left(a^2 x^{\frac{3}{2}}-2 a x^{\frac{5}{2}}+x^{\frac{7}{2}}\right) d x \\
& =a^2 \int_0^a x^{\frac{3}{2}} d x-2 a \int_0^a x^{\frac{5}{2}} d x+\int_0^a x^{\frac{7}{2}} d x \\
& =a^2\left[\frac{x^{\frac{5}{2}}}{\left(\frac{5}{2}\right)}\right]_0^a-2 a\left[\frac{x^{\frac{7}{2}}}{\left(\frac{7}{2}\right)}\right]_0^a+\left[\frac{x^{\frac{9}{2}}}{\left(\frac{9}{2}\right)}\right]_0^a \\
& =\frac{2 a^2}{5}\left[a^{\frac{5}{2}}-0\right]-\frac{4 a}{7}\left[a^{\frac{7}{2}}-0\right]+\frac{2}{9}\left[a^{\frac{9}{2}}-0\right] \\
& =\frac{2}{5} a^{\frac{9}{2}}-\frac{4}{7} a^{\frac{9}{2}}+\frac{2}{9} a^{\frac{9}{2}}=\left(\frac{2}{5}-\frac{4}{7}+\frac{2}{9}\right) a^{\frac{9}{2}} \\
& =\left(\frac{126-180+70}{315}\right) a^{\frac{9}{2}}=\frac{16}{315} a^{\frac{9}{2}} .
\end{aligned}
$

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Question 113 Marks

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

Answer

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

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Question 123 Marks

Evaluate the following definite integrals : If $\int_1^a\left(3 x^2+2 x+1\right) d x=11$, find ' $a$ '.

Answer

$
\begin{aligned}
& \text { Let I }=\int_1^a\left(3 x^2+2 x+1\right) d x \\
& =\left[3\left(\frac{x^3}{3}\right)+2\left(\frac{x^2}{2}\right)+x\right]_1^a \\
& =\left[x^3+x^2+x\right]_1^a \\
& =\left(a^3+a^2+a\right)-(1+1+1) \\
& =a^3+a^2+a-3
\end{aligned}
$
$
\therefore I =11 \text { gives } a ^3+ a ^2+ a -3=11
$
$
\therefore a ^3+ a ^2+ a -14=0
$
$
\therefore\left(a^3-8\right)+\left(a^2+a-6\right)=0
$
$
\therefore(a-2)\left(a^2+2 a+4\right)+(a+3)(a-2)=0
$
$
\therefore(a-2)\left(a^2+2 a+4+a+3\right)=0
$
$
\therefore(a-2)\left(a^2+3 a+7\right)=0
$
$
\therefore a-2=0 \text { or } a ^2+3 a +7=0
$
$
\therefore a =2 \text { or } a =\frac{-3 \pm \sqrt{9-28}}{2}
$
The latter two roots are not real.
$\therefore$ they are rejected.
$
\therefore a =2 \text {. }
$

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Maths (commerce) Solve the Following Question.(3 Marks)

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