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  • Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.1 Question 49 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 49 Maths Textbook Solution.

Answers (1)

Answer:

\int_{0}^{1}\left ( xe^{x}+\cos \frac{\pi x}{4} \right )dx

Sol:

\int_{0}^{1}\left ( xe^{x}+\cos \frac{\pi x}{4} \right )dx=\int_{0}^{1} xe^{x}dx+\int_{0}^{1}\cos \frac{\pi x}{4}dx

Applying integration by parts method in !st integral then

=\left[x \int e^{x} d x\right]_{0}^{1}-\int_{0}^{1}\left(\frac{d(x)}{d x} \int e^{x} d x\right) d x+\left[\frac{\sin \frac{\pi}{4} x}{\frac{\pi}{4}}\right]_{0}^{1}

=\left[x e^{x}\right]_{0}^{1}-\int_{0}^{1} 1 . e^{x} d x+\frac{4}{\pi}\left[\sin \frac{\pi x}{4}\right]_{0}^{1}

=\left[1 . e^{1}-0 . e^{0}\right]-\left[e^{x}\right]_{0}^{1}+\frac{4}{\pi}\left[\sin \frac{\pi}{4}-\sin 0\right]

=[e-0]-\left[e^{1}-e^{0}\right]+\frac{4}{\pi}\left[\sin \frac{\pi}{4}-\sin 0\right]

= \left [ e \right ]-[e-1]+\frac{4}{\pi}\left[\frac{1}{\sqrt{2}}-0\right]

= e-e+1+\frac{4}{\pi} \times \frac{1}{\sqrt{2}}

=\frac{2\sqrt{2}}{\pi }+1

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