Exercise 2.3 - Chapter 2 Integral Calculus I 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

$\mathbf{E} 2.3$
Integrate the following with respect to $\mathrm{x}$.
Question 1.
$e^{x \log a}+e^{a \log a}-e^{n \log x}$
Solution:
$
\begin{aligned}
& e^{x \log a}+e^{a \log a}-e^{n \log x}=e^{\log a^x}+e^{\log a^a}-e^{\log x^n} \\
&=a^x+a^a-x^n \\
& \text { So } \int\left(\mathrm{e}^{x \log a}+\mathrm{e}^{a \log a}-\mathrm{e}^{n \log x}\right) d x \\
&=\int a^x d x+\int a^a d x-\int x^n d x+c \\
&=\frac{a^x}{\log a}+a^a x-\frac{x^{n+1}}{n+1}+c
\end{aligned}
$
Question 2.
$\frac{a^x-e^{x \log b}}{e^{x \log a} b^z}$
Solution:
$
\begin{aligned}
\frac{a^x-e^{x \log b}}{e^{x \log a} b^x} & =\frac{a^x-e^{\log b^x}}{e^{\log a^x} b^x}=\frac{a^x-b^x}{a^x b^x} \\
& =\frac{a^x}{a^x b^x}-\frac{b^x}{a^x b^x}=\frac{1}{b^x}-\frac{1}{a^x} \\
& =b^{-x}-a^{-x}
\end{aligned}
$
So $\begin{aligned} \int \frac{a^x-e^{x \log b}}{e^{x \log a} b^x} d x & =\int b^{-x} d x-\int a^{-x} d x+c \\ & =\frac{b^{-x}}{-\log b}-\frac{a^{-x}}{-\log a}+c\left\{\int a^{-x} d x=\frac{-1}{\log a} a^{-x}\right\} \\ & =\frac{1}{a^x \log a}-\frac{1}{b^x \log b}+c\end{aligned}$

Question 3.
$
\left(e^x+1\right)^2 e^x
$
Solution:
$
\begin{aligned}
\left(e^x+1\right)^2 e^x & =\left(e^{2 x}+2 e^x+1\right) e^x \\
& =e^{3 x}+2 e^{2 x}+e^x
\end{aligned}
$
So $\int\left(e^x+1\right)^2 e^x d x=\int e^{3 x} d x+\int 2 e^{2 x} d x+\int e^x d x$
$
\begin{aligned}
& =\frac{e^{3 x}}{3}+\frac{2 e^{2 x}}{2}+e^x+c \\
& =e^x+e^{2 x}+\frac{e^{3 x}}{3}+c
\end{aligned}
$
Question 4.
$\frac{e^{3 x}-e^{-3 x}}{e^x}$
Solution:
$
\begin{aligned}
& \frac{e^{3 x}-e^{-3 x}}{e^x}=\frac{e^{3 x}}{e^x}-\frac{e^{-3 x}}{e^x} \\
&=e^{2 x}-e^{-4 x} \\
& \operatorname{So} \int \frac{e^{3 x}-e^{-3 x}}{e^x} d x=\int e^{2 x} d x-\int e^{-4 x} d x \\
&=\frac{e^{2 x}}{2}-\frac{e^{-4 x}}{(-4)}+c \\
&=\frac{e^{2 x}}{2}+\frac{e^{-4 x}}{4}+c
\end{aligned}
$

Question 5 .
$\frac{e^{3 x}+e^{5 x}}{e^x+e^{-x}}$
Solution:
$
\begin{aligned}
\frac{e^{3 x}+e^{5 x}}{e^x+e^{-x}} & =\frac{e^{4 x-x}+e^{4 x+x}}{e^x+e^{-x}} \\
& =\frac{e^{4 x}\left(e^{-x}+e^x\right)}{e^x+e^{-x}}=e^{4 x}
\end{aligned}
$
So $\int \frac{e^{3 x}+e^{5 x}}{e^x+e^{-x}} d x=\int e^{4 x} d x$
$
=\frac{e^{4 x}}{4}+c
$
Question 6.
$
\left[1-\frac{1}{x^2}\right] e^{\left(x+\frac{1}{x}\right)}
$
Solution:
$
\begin{aligned}
\int\left[1-\frac{1}{x^2}\right] e^{\left(x+\frac{1}{x}\right)} & =\int e^{\left(x+\frac{1}{x}\right)} d\left(x+\frac{1}{x}\right) \\
& =e^{\left(x+\frac{1}{x}\right)}+c
\end{aligned}
$
Question 7.
$\frac{1}{x(\log x)^2}$
Solution:
$
\begin{aligned}
\int \frac{1}{x(\log x)^2} & =\int \frac{1}{(\log x)^2} d(\log x) \\
& =\int(\log x)^{-2} d(\log x) \\
-(\log x)^{-1}+c & =-\frac{1}{\log x}+c
\end{aligned}
$

Question 8.
If $f^{\prime}(x)=e^x$ and $f(0)=2$, then find $f(x)$.
Solution:
$f^{\prime}(x)=e^x$
Integrating both sides of the equation, $\int f^{\prime}(x) d x=\int e^x d x$
$\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}+\mathrm{c}$
given $f(0)=2$
$2=\mathrm{e}^0+\mathrm{c}$
$\Rightarrow \mathrm{c}=1$
Thus $f(x)=e^x+1$