CBSE Boardहिन्दी माध्यमकक्षा 12 साइन्सगणितसमाकलन3 Marks
Question
मान ज्ञात कीजिए- $\int_0^2 \sqrt{4-x^2} d x$
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Answer
माना $\quad I =\int_0^2 \sqrt{4-x^2} d x$ $\quad$$\quad$$\quad$$=\int_0^2 \sqrt{(2)^2-x^2} d x$ हम जानते हैं $\int \sqrt{a^2-x^2} d x=\frac{1}{2} x \sqrt{a^2-x^2}+\frac{1}{2} a^2 \sin ^{-1}\left(\frac{x}{a}\right)+ C$ इसलिये $\quad I =\left[\frac{1}{2} x \sqrt{4-x^2}+\frac{1}{2} \times 4 \sin ^{-1}\left(\frac{x}{2}\right)\right]_0^2$ $=\left[0+2 \sin ^{-1}(1)-0\right]=2 \sin ^{-1} 1$ $=2 \times \frac{\pi}{2}=\pi$
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