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Maths (commerce) Solve the following Question.(1 Marks)

Integration (p-1) · Maharashtra Board STD 12 Commerce / Arts (MSBSHSE - English Medium). 26 questions.

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Solve the following Question.(1 Marks)

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26 questions · self-marked practice — reveal the answer and mark yourself.

Question 11 Mark
Evalute : $\int \frac{d x}{9 x^2-25}$
Answer
$
\begin{aligned}
& \int \frac{d x}{9 x^2-25}=\frac{1}{9} \int \frac{1}{x^2-\frac{25}{9}} d x \\
= & \frac{1}{9} \int \frac{1}{x^2-\left(\frac{5}{3}\right)^2} d x \\
= & \frac{1}{9} \times \frac{1}{2 \times \frac{5}{3}} \log \left|\frac{x-\frac{5}{3}}{x+\frac{5}{3}}\right|+c \\
= & \frac{1}{30} \log \left|\frac{3 x-5}{3 x+5}\right|+c .
\end{aligned}
$
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Question 21 Mark
Evalute : $\int \frac{d x}{3-2 x-x^2} d x$
Answer
$
\begin{aligned}
& \int \frac{d x}{3-2 x-x^2}=\int \frac{d x}{3-\left(x^2+2 x+1\right)+1} \\
= & \int \frac{1}{(2)^2-(x+1)^2} d x \\
= & \frac{1}{2 \times 2} \log \left|\frac{2+x+1}{2-x-1}\right|+c \\
= & \frac{1}{4} \log \left|\frac{3+x}{1-x}\right|+c .
\end{aligned}
$
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Question 31 Mark
Evalute : $\int \frac{d x}{\sqrt{4 x^2-5}} d x$
Answer
$
\begin{aligned}
& \int \frac{d x}{\sqrt{4 x^2-5}}=\frac{1}{2} \int \frac{1}{\sqrt{x^2-\frac{5}{4}}} d x \\
& =\frac{1}{2} \log \left|x+\sqrt{x^2-\frac{5}{4}}\right|+c .
\end{aligned}
$
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Question 41 Mark
Evalute : $\int \frac{a e^x+b e^{-x}}{\left(a e^x-b e^{-x}\right)^2} d x$
Answer
$
\text { Let } I=\int \frac{a e^x+b e^{-x}}{\left(a e^x-b e^{-x}\right)^2} d x
$
Put $a e^x-b e^{-x}=t$
$
\begin{aligned}
& \therefore\left(a e^x+b e^{-x}\right) d x=d t \\
& \begin{aligned}
\therefore I & =\int \frac{1}{t^2} d t=\int t^{-2} d t \\
& =\frac{t^{-1}}{-1}+c=\frac{-1}{t}+c \\
& =\frac{-1}{a e^x+b e^{-x}}+c .
\end{aligned}
\end{aligned}
$
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Question 51 Mark
Evalute : $\int|x| d x$ if $x<0$
Answer
$
\begin{aligned}
& \int|x| d x=\int-x d x \ldots .[\because x<0] \\
& =-\int x d x \\
& =-\frac{x^2}{2}+c
\end{aligned}
$
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Question 61 Mark
Evalute : If $f^{\prime}(x)=\sqrt{x}$ and $f(1)=2$, then find the value of $f(x)$.
Answer
By the definition of integral
$
\begin{gathered}
f(x)=\int f^{\prime}(x) d x=\int \sqrt{x} d x \\
=\frac{x^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+c=\frac{2}{3} x^{\frac{3}{2}}+c \\
\therefore f(1)=\frac{2}{3}(1)^{\frac{3}{2}}+c=\frac{2}{3}+c
\end{gathered}
$
But $f(1)=2$
$
\begin{aligned}
& \therefore \frac{2}{3}+c=2 \quad \therefore c=\frac{4}{3} \\
& \therefore \text { from (1), } f(x)=\frac{2}{3} x^{\frac{3}{2}}+\frac{4}{3} .
\end{aligned}
$
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Question 71 Mark
Evalute : $\int \frac{x-1}{\sqrt{x+4}} d x$
Answer
$
\int \frac{x-1}{\sqrt{x+4}} d x=\int \frac{(x+4)-5}{\sqrt{x+4}} d x
$
$
\begin{aligned}
& =\int\left(\frac{x+4}{\sqrt{x+4}}-\frac{5}{\sqrt{x+4}}\right) d x \\
& =\int\left(\sqrt{x+4}-\frac{5}{\sqrt{x+4}}\right) d x \\
& =\int(x+4)^{\frac{1}{2}} d x-5 \int(x+4)^{-\frac{1}{2}} d x \\
& =\frac{(x+4)^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}-5 \cdot \frac{(x+4)^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}+c
\end{aligned}
$
$2{ }_3^3$ $=\frac{2}{3}(x+4)^{\frac{3}{2}}-10 \sqrt{x+4}+c$.
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Question 81 Mark
Evalute : $\int \frac{1}{2 x+3} d x$
Answer
$
\begin{aligned}
& \quad \int \frac{1}{2 x+3} d x=\frac{\log |2 x+3|}{2}+c \\
& =\frac{1}{2} \log |2 x+3|+c .
\end{aligned}
$
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Question 91 Mark
Evalute : $\int(5 x+1)^{\frac{4}{9}} d x$
Answer
$
\begin{aligned}
\int(5 x+1)^{\frac{4}{9}} d x & =\frac{(5 x+1)^{\frac{4}{9}+1}}{\frac{4}{9}+1} \times \frac{1}{5}+c \\
& =\frac{9}{65}(5 x+1)^{\frac{13}{9}}+c .
\end{aligned}
$
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Question 101 Mark
Evalute : $\int e^x\left[(\log x)^2+\frac{2 \log x}{x}\right] d x$
Answer
$
\text { Let } I=\int e^x\left[(\log x)^2+\frac{2 \log x}{x}\right] d x
$
Put $f(x)=(\log x)^2$
Then $f^{\prime}(x)=\frac{d}{d x}(\log x)^2=2 \log x \cdot \frac{d}{d x}(\log x)$
$
\begin{aligned}
& =2 \log x \times \frac{1}{x}=\frac{2 \log x}{x} \\
\therefore I & =\int e^x\left[f(x)+f^{\prime}(x)\right] d x \\
& = e ^x \cdot f(x)+c=e^x \cdot(\log x)^2+c
\end{aligned}
$
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Question 111 Mark
Evalute : $\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x$
Answer
Let $I=\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x$
Put $f(x)=\frac{1}{x}$. Then $f^{\prime}(x)=-\frac{1}{x^2}$
$
\begin{aligned}
\therefore I & =\int e^x-\left[f(x)+f^{\prime}(x)\right] d x \\
& =e^x \cdot f(x)+c= e ^x \cdot \frac{1}{x}+c .
\end{aligned}
$
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Question 121 Mark
Evalute : $\int x \log x$
Answer
$
\begin{aligned}
& \int x \log x d x=\int(\log x) \cdot x d x \\
= & (\log x) \int x d x-\int\left[\frac{d}{d x}(\log x) \int x d x\right] d x \\
= & (\log x) \cdot \frac{x^2}{2}-\int \frac{1}{x} \cdot \frac{x^2}{2} d x \\
= & \frac{1}{2} x^2 \log x-\frac{1}{2} \int x d x \\
= & \frac{x^2}{2} \log x-\frac{1}{2} \cdot \frac{x^2}{2}+c=\frac{x^2}{2} \log x-\frac{x^2}{4}+c .
\end{aligned}
$
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Question 131 Mark
Evalute : $\int \frac{1}{\sqrt{x^2+4 x+29}} d x$
Answer
$
\begin{aligned}
& \int \frac{1}{\sqrt{x^2+4 x+29}} d x \\
= & \int \frac{1}{\sqrt{\left(x^2+4 x+4\right)+25}} d x \\
= & \int \frac{1}{\sqrt{(x+2)^2+(5)^2}} d x \\
= & \log \left|(x+2)+\sqrt{(x+2)^2+(5)^2}\right|+c
\end{aligned}
$
$
=\log \left|(x+2)+\sqrt{x^2+4 x+29}\right|+c \text {. }
$
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Question 141 Mark
Evalute : $\int \frac{1}{\sqrt{3 x^2+8}} d x$
Answer
$
\begin{aligned}
& \int \frac{1}{\sqrt{3 x^2+8}} d x=\int \frac{1}{\sqrt{(\sqrt{3} x)^2+(\sqrt{8})^2}} d x \\
= & \frac{\log \left|\sqrt{3} x+\sqrt{(\sqrt{3} x)^2+(\sqrt{8})^2}\right|}{\sqrt{3}}+c \\
= & \frac{1}{\sqrt{3}} \log \left|\sqrt{3} x+\sqrt{3 x^2+8}\right|+c .
\end{aligned}
$
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Question 151 Mark
Evalute : $\int \frac{1}{7+6 x-x^2} d x$
Answer
$
\begin{aligned}
& \int \frac{1}{7+6 x-x^2} d x \\
= & \int \frac{1}{7-\left(x^2-6 x+9\right)+9} d x \\
= & \int \frac{1}{(4)^2-(x-3)^2} d x \\
= & \frac{1}{2 \times 4} \log \left|\frac{4+x-3}{4-x+3}\right|+c \\
= & \frac{1}{8} \log \left|\frac{1+x}{7-x}\right|+c .
\end{aligned}
$
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Question 161 Mark
Evalute : $\int \frac{1}{\sqrt{x^2-8 x-20}} d x$
Answer
$
\begin{aligned}
& \int \frac{1}{\sqrt{x^2-8 x-20}} d x \\
= & \int \frac{1}{\sqrt{\left(x^2-8 x+16\right)-16-20}} d x \\
= & \int \frac{1}{\sqrt{(x-4)^2-(6)^2}} d x \\
= & \log \left|(x-4)+\sqrt{(x-4)^2-(6)^2}\right|+c \\
= & \log \left|(x-4)+\sqrt{x^2-8 x-20}\right|+c .
\end{aligned}
$
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Question 171 Mark
Evalute : $\int \frac{1}{\sqrt{3 x^2-5}} d x$
Answer
$
\begin{aligned}
& \int \frac{1}{\sqrt{3 x^2-5}} d x=\int \frac{1}{\sqrt{(\sqrt{3} x)^2-(\sqrt{5})^2}} d x \\
= & \frac{\log \left|\sqrt{3} x+\sqrt{(\sqrt{3} x)^2-(\sqrt{5})^2}\right|}{\sqrt{3}}+c \\
= & \frac{1}{\sqrt{3}} \log \left|\sqrt{3} x+\sqrt{3 x^2-5}\right|+c .
\end{aligned}
$
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Question 181 Mark
Evalute : $\int \frac{2 x+6}{\sqrt{x^2+6 x+3}} d x$
Answer
$
\text { Let } I=\int \frac{2 x+6}{\sqrt{x^2+6 x+3}} d x
$
Put $x^2+6 x+3=t$
$
\begin{aligned}
& \therefore(2 x+6) d x=d t \\
& \therefore I=\int \frac{1}{\sqrt{t}} d t=\int t^{-\frac{1}{2}} d t
\end{aligned}
$
$
=\frac{t^{\frac{1}{2}}}{1 / 2}+c=2 \sqrt{x^2+6 x+3}+c .
$
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Question 191 Mark
Evalute : $\int \frac{1}{x \log x} d x$
Answer
Put $\log x=t$
$
\begin{aligned}
& \therefore \frac{1}{x} dx = dt \\
& \therefore \int \frac{d x}{x \cdot \log x}=\int \frac{1}{\log x} \cdot \frac{1}{x} d x \\
& =\int \frac{1}{t} dt \\
& =\log | t |+ C \\
& =\log |\log x |+ c
\end{aligned}
$
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Question 201 Mark
Evalute : $\int(x+1)(x+2)^7(x+3) d x$
Answer
$
\begin{aligned}
& \text { Let I }=\int(x+1)(x+2)^7(x+3) d x \\
& =\int(x+2)^7(x+1)(x+3) d x \\
& =\int(x+2)^7[(x+2)-1][(x+2)+1] d x \\
& =\int(x+2)^7\left[(x+2)^2-1\right] d x \\
& =\int\left[(x+2)^9-(x+2)^7\right] d x \\
& =\int(x+2)^9 d x-\int(x+2)^7 d x \\
& =\frac{(x+2)^{10}}{10}-\frac{(x+2)^8}{8}+c
\end{aligned}
$
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Question 211 Mark
Evalute : $\int\left(e^x+e^{-x}\right)^2\left(e^x-e^{-x}\right) d x$
Answer
Let $I=\int\left(e^x+e^{-x}\right)^2\left(e^x-e^{-x}\right) d x$
Put $e^x+e^{-x}=t$
$
\begin{aligned}
& \therefore\left(e^x-e^{-x}\right) d x=d t \\
& \begin{aligned}
\therefore I & =\int t^2 d t=\frac{t^3}{3}+c \\
& =\frac{\left(e^x+e^{-x}\right)^3}{3}+c .
\end{aligned}
\end{aligned}
$
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Question 221 Mark
If $f^{\prime}(x)=x^2+5$ and $f(0)=-1$, then find the value of $f(x)$.
Answer
By the definition of integral
$
\begin{aligned}
& f(x)=\int f^{\prime}(x) d x \\
& =\int\left(x^2+5\right) d x \\
& =\int x^2 d x+5 \int 1 d x \\
& =\frac{x^3}{3}+5 x+C
\end{aligned}
$
Now, $f(0)=-1$ gives
$
\begin{aligned}
& f(0)=0+0+c=-1 \\
& \therefore c=-1
\end{aligned}
$
$\therefore$ from (1), $f(x)=\frac{x^3}{3}+5 x-1$.
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Question 231 Mark
Evaluate $\int \frac{1}{x(x-1)} d x$
Answer
$
\begin{aligned}
& \int \frac{1}{x(x-1)} d x=\int \frac{x-(x-1)}{x(x-1)} d x \\
= & \int\left(\frac{1}{x-1}-\frac{1}{x}\right) d x \\
= & \int \frac{1}{x-1} d x-\int \frac{1}{x} d x \\
= & \log |x-1|-\log |x|+c \\
= & \log \left|\frac{x-1}{x}\right|+c .
\end{aligned}
$
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Question 241 Mark
Evaluate $\int\left(3 x^2-5\right)^2 d x$
Answer
$
\begin{aligned}
& \int\left(3 x^2-5\right)^2 d x \\
& =\int\left(9 x^4-30 x^2+25\right) d x \\
& =9 \int x^4 d x-30 \int x^2 d x+25 \int 1 d x \\
& =9\left(\frac{x^5}{5}\right)-30\left(\frac{x^3}{3}\right)+25 x+c \\
& =\frac{9 x^5}{5}-10 x^3+25 x+c
\end{aligned}
$
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Question 251 Mark
Evaluate $\int \frac{3 x^3-2 \sqrt{x}}{x} d x$
Answer
$
\begin{aligned}
& \int \frac{3 x^3-2 \sqrt{x}}{x} d x \\
= & \int\left(\frac{3 x^3}{x}-\frac{2 \sqrt{x}}{x}\right) d x \\
= & \int\left(3 x^2-\frac{2}{\sqrt{x}}\right) d x \\
= & 3 \int x^2 d x-2 \int x^{-\frac{1}{2}} d x
\end{aligned}
$
$
\begin{aligned}
& =3 \cdot\left(\frac{x^3}{3}\right)-2 \cdot\left[\frac{x^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}\right]+c \\
& =x^3-4 \sqrt{x}+c .
\end{aligned}
$
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Question 261 Mark
Evaluate $\int\left(1+x+\frac{x^2}{2 !}\right) d x$
Answer
$
\int\left(1+x+\frac{x^2}{2 !}\right) d x
$
$
\begin{aligned}
& =\int 1 d x+\int x d x+\frac{1}{2 !} \int x^2 d x \\
& =x+\frac{x^2}{2}+\frac{1}{2 !} \times \frac{x^3}{3}+c \\
& =x+\frac{x^2}{2}+\frac{x^3}{6}+c .
\end{aligned}
$
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