मराठी

Π / 4 ∫ 0 E X Sin X D X

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प्रश्न

\[\int\limits_0^{\pi/4} e^x \sin x dx\]

बेरीज
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उत्तर

\[Let, I = \int_0^\frac{\pi}{4} e^x \sin x d x ..............(1)\]
\[ = \left[ - e^x \cos x \right]_0^\frac{\pi}{4} + \int_0^\frac{\pi}{4} e^x \cos x dx\]
\[ = \left[ - e^x \cos x \right]_0^\frac{\pi}{4} + \left[ e^x \sin x \right]_0^\frac{\pi}{4} - \int_0^\frac{\pi}{4} e^x \sin x dx\]
\[ \Rightarrow I = \left[ - e^x \cos x \right]_0^\frac{\pi}{4} + \left[ e^x \sin x \right]_0^\frac{\pi}{4} - I ..............\left[\text{Using (1)} \right] \]
\[ \Rightarrow 2I = \left[ - e^x \cos x \right]_0^\frac{\pi}{4} + \left[ e^x \sin x \right]_0^\frac{\pi}{4} \]
\[ = - \frac{1}{\sqrt{2}} e^\frac{\pi}{4} + 1 + \frac{1}{\sqrt{2}} e^\frac{\pi}{4} - 0\]
\[ = 1\]
\[\text{Hence }I = \frac{1}{2}\]

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पाठ 19: Definite Integrals - Revision Exercise [पृष्ठ १२२]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Revision Exercise | Q 27 | पृष्ठ १२२

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