Exercise 2.1 - Chapter 2 Integral Calculus I 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Integral Calculus I
$\operatorname{Ex} 2.1$
Integrate the following with respect to $\mathrm{x}$.
Question 1.
$\sqrt{3 x+5}$
Solution:
$\sqrt{3 x+5}=(3 x+5)^{\frac{1}{2}}$
This is of the form $(a x+b)^n$ where $a=3, b=5, n=1 / 2$
$
\int(a x+b)^n d x=\frac{(a x+b)^{n+1}}{a(n+1)}+c
$
So $\int(3 x+5)^{\frac{1}{2}} d x=\frac{(3 x+5)^{\frac{1}{2}+1}}{3\left(\frac{1}{2}+1\right)}+c$
$
\begin{aligned}
& =\frac{(3 x+5)^{\frac{3}{2}}}{3\left(\frac{3}{2}\right)}=\frac{(3 x+5)^{\frac{3}{2}}}{\frac{9}{2}}+c \\
& =\frac{2}{9}(3 x+5)^{\frac{3}{2}}+c
\end{aligned}
$
Question 2.
$
\left(9 x^2-\frac{4}{x^2}\right)^2
$
Solution:

$
\begin{aligned}
\left(9 x^2-\frac{4}{x^2}\right)^2 & =\left(9 x^2\right)^2-2\left(9 x^2\right)\left(\frac{4}{x^2}\right)+\left(\frac{4}{x^2}\right)^2 \\
& =81 x^4-72+\frac{16}{x^4}
\end{aligned}
$
So $\int\left(9 x^2-\frac{4}{x^2}\right)^2 d x=\int 81 x^4 d x-\int 72 d x+\int \frac{16}{x^4} d x+c$
$
\begin{aligned}
& =\frac{81 x^5}{5}-72 x+16 \int x^{-4} d x+c \\
& =\frac{81 x^5}{5}-72 x+16 \frac{(x)^{-3}}{(-3)}+c=\frac{81 x^5}{5}-72 x-\frac{16}{3 x^3}+c
\end{aligned}
$
Question 3.
$
(3+x)(2-5 x)
$
Solution:
$
\begin{aligned}
(3+x)(2-5 x) & =6-15 x+2 x-5 x^2 \\
& =6-13 x-5 x^2
\end{aligned}
$
$
\text { So } \begin{aligned}
\int(3+x)(2-5 x) d x & =\int\left(6-13 x-5 x^2\right) d x \\
& =\int 6 d x-\int 13 x d x-\int 5 x^2 d x+c \\
& =6 x-\frac{13 x^2}{2}-\frac{5 x^3}{3}+c
\end{aligned}
$
Question 4.
$
\sqrt{ } x\left(x^3-2 x+3\right)
$

Solution:
$
\begin{aligned}
\sqrt{x}\left(x^3-2 x+3\right) & =x^{\frac{1}{2}}\left(x^3-2 x+3\right) \\
& =x^{\frac{7}{2}}-2 x^{\frac{3}{2}}+3 x^{\frac{1}{2}}
\end{aligned}
$
So
$
\begin{aligned}
\int \sqrt{x}\left(x^3-2 x+3\right) d x & =\int x^{\frac{7}{2}} d x-\int 2 x^{\frac{3}{2}} d x+\int 3 x^{\frac{1}{2}} d x \\
& =\frac{x^{\frac{7}{2}+1}}{\frac{7}{2}+1}-\frac{2 x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{3 x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c \\
& =\frac{2}{9} x^{\frac{9}{2}}-\frac{2 x^{\frac{5}{2}}}{\frac{5}{2}}+\frac{3 x^{\frac{3}{2}}}{\frac{3}{2}}+c \\
& =\frac{2}{9} x^{\frac{9}{2}}-\frac{4}{5} x^{\frac{5}{2}}+2 x^{\frac{3}{2}}+c
\end{aligned}
$

Question 5.
$\frac{8 x+13}{\sqrt{4 x+7}}$
Solution:
$
\begin{aligned}
\frac{8 x+13}{\sqrt{4 x+7}} & =\frac{8 x+14-1}{\sqrt{4 x+7}} \\
& =\frac{2(4 x+7)-1}{\sqrt{4 x+7}} \\
& =\frac{2(4 x+7)}{\sqrt{4 x+7}}-\frac{1}{\sqrt{4 x+7}} \\
& =2(4 x+7)^{\frac{1}{2}}-(4 x+7)^{-\frac{1}{2}}
\end{aligned}
$
So $\int \frac{8 x+13}{\sqrt{4 x+7}}=2 \int(4 x+7)^{\frac{1}{2}} d x-\int(4 x+7)^{-\frac{1}{2}} d x+c$
$
\begin{aligned}
& =\frac{2(4 x+7)^{\frac{3}{2}}}{4\left(\frac{3}{2}\right)}-\frac{(4 x+7)^{\frac{1}{2}}}{4\left(\frac{1}{2}\right)}+c \\
& =\frac{(4 x+7)^{\frac{3}{2}}}{3}-\frac{(4 x+7)^{\frac{1}{2}}}{2}+c
\end{aligned}
$

Question 6.
$
\frac{1}{\sqrt{x+1}+\sqrt{x-1}}
$
Solution:
$
\begin{aligned}
\frac{1}{\sqrt{x+1}+\sqrt{x-1}} & =\frac{1}{\sqrt{x+1}+\sqrt{x-1}} \times \frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} \\
& =\frac{\sqrt{x+1}-\sqrt{x-1}}{x+1-(x-1)}=\frac{\sqrt{x+1}-\sqrt{x-1}}{2} \\
& =\frac{(x+1)^{\frac{1}{2}}}{2} \frac{(x-1)^{\frac{1}{2}}}{2}
\end{aligned}
$
So
$
\begin{aligned}
\frac{1}{\sqrt{x+1}+\sqrt{x-1}} & =\frac{1}{2} \int(x+1)^{\frac{1}{2}} d x-\frac{1}{2} \int(x-1)^{\frac{1}{2}} d x+c \\
& =\frac{1}{2} \frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{2} \frac{(x-1)^{\frac{3}{2}}}{\frac{3}{2}}+c \\
& =\frac{1}{3}\left[(x+1)^{\frac{3}{2}}-(x-1)^{\frac{3}{2}}\right]+c
\end{aligned}
$

Question 7.
If $f^{\prime}(x)=x+b, f(1)=5$ and $f(2)=13$, then find $f(x)$
Solution:
$
f^{\prime}(x)=x+b
$
Integrating both sides of the equations
$
\begin{gathered}
\int f^{\prime}(x) d x=\int x d x+\int b d x+c \\
\Rightarrow \quad f(x)=\frac{x^2}{2}+b x+c \\
f(1)=5 \Rightarrow 5=\frac{1}{2}+b+c \Rightarrow b+c=\frac{9}{2} \\
f(2)=13 \Rightarrow 13=\frac{4}{2}+2 b+c \Rightarrow 2 b+c=11 \\
(2)-(1) \text { gives } b=\frac{13}{2}, c=\frac{9}{2}-\frac{13}{2}=-2 \\
\text { Thus } \quad f(x)=\frac{x^2}{2}+\frac{13}{2} x-2
\end{gathered}
$

Question 8.
If $f(x)=8 x^3-2 x$ and $f(2)=8$, then find $f(x)$
Solution:
$
f^{\prime}(x)=8 x^3-2 x
$
Integrating both sides of the equation,
$
\begin{aligned}
\int f^{\prime}(x) d x & =\int 8 x^3 d x-\int 2 x d x+c \\
f(x) & =\frac{8 x^4}{4}-\frac{2 x^2}{2}+c \\
f(x) & =2 x^4-x^2+c \\
f(2) & =8 \\
8 & =2(2)^4-(2)^2+c \\
8 & =32-4+c \\
c & =-20
\end{aligned}
$
Thus $\quad f(x)=2 x^4-x^2-20$
Type: II
(i) $\int \frac{1}{x} d x=\log |x|+c$
(ii) $\int \frac{1}{a x+b} d x=\frac{1}{a} \log |a x+b|+c$

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