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  • Please Solve R.D.Sharma class 12 Chapter 19 Definite Integrals Exercise 19.1 Question 10 Maths textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 19 Definite Integrals Exercise 19.1 Question 10 Maths textbook Solution.

Answers (1)

Answer: 2

Hint: Use indefinite formula to solve the integral and then put the value of limit to get the required answer

Given: \int_{0}^{\frac{\pi }{2}}\left ( \sin x+\cos x \right )dx

Solution:

\int_{0}^{\frac{\pi }{2}}\left ( \sin x+\cos x \right )dx=\int_{0}^{\frac{\pi }{2}}\sin xdx+\int_{0}^{\frac{\pi }{2}}\cos \: xdx

=[-\cos x]_{0}^{\frac{\pi}{2}}+[\sin x]_{0}^{\frac{\pi}{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \int \sin x d x=-\cos x \\ \int \cos x d x=\sin x \end{array}\right]

=-\left[\cos \frac{\pi}{2}-\cos 0\right]+\left[\sin \frac{\pi}{2}-\sin 0\right]\left[\begin{array}{l} \cos \frac{\pi}{2}=0, \cos 0=1 \\ \sin \frac{\pi}{2}=1, \sin 0=0 \end{array}\right]

=-\left ( 0-1 \right )+\left ( 1-0 \right )

=-\left ( -1 \right )+1

=1+1=2

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