Exercise 2.5 - Chapter 2 Integral Calculus I 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $2.5$
Integrate the following with respect to $\mathrm{x}$.
Question 1.
$\mathrm{x} \mathrm{e}^{-\mathrm{x}}$
Solution:

$\begin{aligned}
\int x e^{-x} d x & =\int u d v \\
\text { Let } u=x \text { and } d v & =e^{-x} d x \\
\text { then } d u=d x, v & =-e^{-x} \\
u v-\int v d u & =-x e^{-x}+\int e^{-x} d x \\
& =-x e^{-x}-e^{-x}+c \\
& =-e^{-x}(x+1)+c
\end{aligned}$

Question 2.
$
x^3 e^{3 x}
$
Solution:
$
\begin{array}{rlrl}
\int x^3 e^{3 x} d x & =\int u d v & & \\
& =u v-u^{\prime} v_1+u^{\prime \prime} v_2-u^{\prime \prime \prime} v_3+\ldots & \\
u & =x^3 & d \nu & =e^{3 x} d x \\
u^{\prime} & =3 x^2 & v & =\frac{e^{3 x}}{3} \\
u^{\prime \prime} & =6 x & v_1 & =\frac{e^{3 x}}{9} \\
u^{\prime \prime \prime} & =6 & v_2 & =\frac{e^{3 x}}{27} \\
v_3 & =\frac{e^{3 x}}{81}
\end{array}
$
So
$
\begin{aligned}
\int x^3 e^{3 x} d x & =x^3\left(\frac{e^{3 x}}{3}\right)-\left(3 x^2\right)\left(\frac{e^{3 x}}{9}\right)+6 x\left(\frac{e^{3 x}}{27}\right)-6\left(\frac{e^{3 x}}{81}\right)+c \\
& =e^{3 x}\left[\frac{x^3}{3}-\frac{x^2}{3}+\frac{2 x}{9}-\frac{2}{27}\right]+c
\end{aligned}
$

Question 3.
$\log \mathrm{x}$
Solution:
$
\begin{aligned}
\int \log x d x & =\int u d v \\
\text { Where } u=\log x \Rightarrow d u & =\frac{1}{x} \\
d v=d x \Rightarrow v & =x \\
x \log x-\int x\left(\frac{1}{x}\right) d x & =x \log x-x+c \\
& =x(\log x-1)+c
\end{aligned}
$
Question 4.
$\mathrm{x} \log \mathrm{x}$
Solution:
$
\int x \log x d x=\int u d v \quad \text { where } u=\log x \text { and } d v=x d x
$
So $\quad d u=\frac{1}{x} d x$ and $v=\frac{x^2}{2}$
$
\begin{aligned}
\Rightarrow \quad \int x \log x d x & =\frac{x^2}{2} \log x-\int \frac{x^2}{2}\left(\frac{1}{x}\right) d x \\
& =\frac{x^2}{2} \log x-\frac{1}{2} \int x d x \\
& =\frac{x^2}{2} \log x-\frac{x^2}{4}+\mathrm{c} \\
& =\frac{x^2}{2}\left[\log x-\frac{1}{2}\right]+c
\end{aligned}
$

Question 5.
$\mathrm{x}^{\mathrm{n}} \log \mathrm{x}$
Solution:
$
x^n \log x d x=\int u d v=u v-\int v d u
$
where $u=\log x$ and $d v=x^n d x$
$
\begin{aligned}
\Rightarrow d u=\frac{1}{x} & \left(v=\frac{x^{n+1}}{n+1}\right) \\
& =\frac{x^{n+1}}{n+1} \log x-\int \frac{x^{n+1}}{n+1}\left(\frac{1}{x}\right) \\
& =\frac{x^{n+1}}{n+1} \log x-\frac{1}{n+1} \int x^n d x \\
& =\frac{x^{n+1}}{n+1} \log x-\frac{x^{n+1}}{(n+1)(n+1)}+c \\
& =\frac{x^{n+1}}{n+1}\left[\log x-\frac{1}{n+1}\right]+c
\end{aligned}
$

Question 6.
$
x^5 e^{x^2}
$
Solution:
Consider $\int 2 x e^{x^2} d x=\int e^{x^2} d\left(x^2\right)$
$
=e^{x^2}+k \text {. }
$
We use this result to evaluate the integral
$
\begin{aligned}
\int x^5 e^{x^2} d x & =\frac{2 x x^4 e^{x^2} d x}{2} \\
& =\frac{1}{2} \int x^4\left(2 x e^{x^2}\right) d x
\end{aligned}
$
Let $u=x^4 \Rightarrow d u=4 x^3 d x$
$
d v=2 x e^{x^2} d x
$
$
v=e^{x^2}
$
(using (A))
So
$
\begin{aligned}
\int x^5 e^{x^2} d x & =\frac{1}{2}\left[x^4 e^{x^2}-4 \int x^3 e^{x^2} d x\right] \\
& =\frac{x^4 e^{x^2}}{2}-\int x^2\left(2 x e^{x^2}\right) d x \ldots
\end{aligned}
$
Consider $\int x^2\left(2 x e^{x^2}\right) d x$
Let $u=x^2 \Rightarrow d u=2 x d x$
$
\begin{aligned}
& d v=2 x e^{x^2} d x \\
& v=e^{x^2} \quad=x^2 e^{x^2}-\int 2 x e^{x^2} d x \\
& =x^2 e^{x^2}-e^{x^2} \\
&
\end{aligned}
$
Using (2) in (1) we get
$
\begin{aligned}
\int x^5 e^{x^2} d x & =\frac{x^4 e^{x^2}}{2}-x^2 e^{x^2}+e^{x^2}+c \\
& =\frac{e^{x^2}}{2}\left(x^4-2 x^2+2\right)+c
\end{aligned}
$

$
\begin{aligned}
\int x^5 e^{x^2} d x \text { Let } x^2=t \quad & \therefore 2 x d x=d t \\
& =\int x^4\left(x e^{x^2}\right) d x \\
& =\frac{1}{2} \int t^2 e^t d t
\end{aligned}
$
Now we use integration by parts
$
\begin{array}{rlrl}
u & =t^2 & d v & =e^t d t \\
u^{\prime}=2 t & u^{\prime \prime}=2 & & e^t \\
u^{\prime \prime \prime}=0 & v_1 & =e^t \\
v_2 & =e^t \\
& \int t^2 e^t d t & =t^2 e^t-2 t e^t+2 e^t
\end{array}
$
Using $t^2=x^4$ and $e^t=e^{x^2}$ we get
$
\begin{aligned}
\int x^5 e^{x^2} d x & =\frac{1}{2}\left(x^4 e^{x^2}-2 x^2 e^{x^2}+2 e^{x^2}\right)+c \\
& =\frac{e^{x^2}}{2}\left(x^4-2 x^2+2\right)+c
\end{aligned}
$