Solve the Following Question.(4 Marks)

Question 14 Marks

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

Answer

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

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Question 24 Marks

Evaluate the following definite integrals : $\int_2^3 \frac{x}{(x+2)(x+3)} d x$

Answer

$
\text { Let } I =\int_2^3 \frac{x}{(x+2)(x+3)} d x
$
Let $\frac{x}{(x+2)(x+3)}=\frac{A}{x+3}+\frac{B}{x+2}$
$
\therefore x = A ( x +2)+ B ( x +3)
$
Put $x+3=0$, i.e. $x=-3$, we get
$
\begin{aligned}
& -3=A(-1)+B(0) \\
& \therefore A=3
\end{aligned}
$
Put $x+2=0$, i.e. $x=-2$, we get
$
\begin{aligned}
& -2= A (0)+ B (1) \\
& \therefore B =-2
\end{aligned}
$
$
\therefore \frac{x}{(x+2)(x+3)}=\frac{3}{x+3}+\frac{(-2)}{x+2}
$
$
\begin{aligned}
\therefore I & =\int_2^3\left[\frac{3}{x+3}+\frac{(-2)}{x+2}\right] d x \\
& =[3 \log (x+3)-2 \log (x+2)]_2^3 \\
& =[3 \log (3+3)-2 \log (3+2)]-
\end{aligned}
$
$[3 \log (2+3)-2 \log (2+2)]$
$
\begin{gathered}
=3 \log 6-5 \log 5+2 \log 4 \\
=\log 6^3-\log 5^5+\log 4^2 \\
=\log 216-\log 3125+\log 16 \\
=\log \left(\frac{216 \times 16}{3125}\right)=\log \left(\frac{3456}{3125}\right) .
\end{gathered}
$

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Maths (commerce) Solve the Following Question.(4 Marks)

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