निश्चित समाकलन intlimits_0^{sqrt {ln left( {f | vedclass

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(C) माना $t = e^{x^2}$.तब,$dt = e^{x^2} \cdot 2x dx$.जब $x = 0$,तब $t = e^{0^2} = e^0 = 1$.जब $x = \sqrt{\ln(\frac{\pi}{2})}$,तब $t = e^{(\sqrt{\ln(\f

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$1 - \sin(1)$

Vedclass · Answer

(C) माना $t = e^{x^2}$.तब,$dt = e^{x^2} \cdot 2x dx$.जब $x = 0$,तब $t = e^{0^2} = e^0 = 1$.जब $x = \sqrt{\ln(\frac{\pi}{2})}$,तब $t = e^{(\sqrt{\ln(\frac{\pi}{2})})^2} = e^{\ln(\frac{\pi}{2})} = \frac{\pi}{2}$.अतः समाकलन $\int_{1}^{\frac{\pi}{2}} \cos(t) dt$ हो जाता है।समाकलन का मान ज्ञात करने पर: $[\sin(t)]_{1}^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}) - \sin(1) = 1 - \sin(1)$.

निश्चित समाकलन $\int\limits_0^{\sqrt {\ln \left( {\frac{\pi }{2}} \right)} } {\cos \left( {{e^{{x^2}}}} \right)} \cdot 2x {e^{{x^2}}}dx$ का मान ज्ञात कीजिए।

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