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

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

Answers (1)

Answer:

\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}

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