Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 48 Maths Textbook Solution.

$\frac{e^{2}}{4}+\frac{1}{4}+\frac{2}{\pi }$

Hint: Use indefinite formula then put the limits to solve this integral

Given:

$\int_{0}^{1}\left(x e^{2 x}+\sin \frac{\pi x}{2}\right) d x$

Sol:

$\int_{0}^{1}\left(x e^{2 x}+\sin \frac{\pi x}{2}\right) d x=\int_{0}^{1} x e^{2 x} d x+\int_{0}^{1} \sin \frac{\pi x}{2} d x$

Applying integration by parts method in 1st integral then

$\left.=\left[x \int e^{2 x} d x\right]_{0}^{1}-\int_{0}^{1}\left(\frac{d(x)}{d x} \int e^{2 x} d x\right) d x+\left[\frac{\cos \frac{\pi}{2} x}{\frac{\pi}{2}}\right]_{0}^{1}\right]_{0}^{1}$

$=\left[x \frac{e^{2 x}}{2}\right]_{0}^{1}-\int_{0}^{1} \frac{e^{2 x}}{2} d x+\frac{2}{\pi}\left[\cos \frac{\pi x}{2}\right]_{0}^{1}$

$=\left[\frac{1 \cdot e^{2.1}}{2}-\frac{0 . e^{20}}{2}\right]_{0}^{1}-\frac{1}{2} \int_{0}^{1} \frac{e^{2 x}}{2} d x+\frac{2}{\pi}\left[\cos \frac{\pi \times 1}{2}-\cos \frac{\pi \times 0}{2}\right]$

$=\left[\frac{e^{2}}{2}-0\right]-\frac{1}{2}\left[\frac{e^{2 x}}{2}\right]_{0}^{1}+\frac{2}{\pi}\left[\cos \frac{\pi}{2}-\cos 0\right]$

$=\left[\frac{e^{2}}{2}\right]-\frac{1}{4}\left[e^{2 \cdot 1}-e^{2.0}\right]+\frac{2}{\pi}[0-1]$

$=\left[\frac{e^{2}}{2}\right]-\frac{1}{4}\left[e^{2}-1\right]+\frac{2}{\pi}[-1]$

$=\left[\frac{e^{2}}{2}\right]-\frac{e^{2}}{4}+\frac{1}{4}-\frac{2}{\pi}$

$=\frac{2 e^{2}-e^{2}}{4}+\frac{1}{4}+\frac{2}{\pi}$

$=\frac{e^{2}}{4}+\frac{1}{4}+\frac{2}{\pi}$