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  • Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 24 maths

Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 24 maths

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best_answer

Answer: \pi

Hint: you must know the rule of integration

Given:\int_{0}^{2} \sqrt{4-x^{2}}

Solution:  \int_{0}^{2} \sqrt{4-x^{2}}

\begin{aligned} &\text { Put } x=2 \sin \theta \\\\ &d x=2 \cos \theta d \theta \end{aligned}

       \begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \sqrt{4-4 \sin ^{2} \theta} \cdot 2 \cos \theta d \theta \\\\ &=\int_{0}^{\frac{\pi}{2}} 4 \cos ^{2} \theta d \theta \end{aligned}

       \begin{aligned} &=\int_{0}^{\frac{\pi}{2}} 4 \cdot \frac{1}{2}(1+\cos 2 \theta) d \theta \\\\ &=2\left[\theta+\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{2}} \end{aligned}

       \begin{aligned} &=2\left[\frac{\pi}{2}+\frac{\sin \pi}{2}-0-0\right] \\\\ &=\pi \end{aligned}

 

 

 

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