Tích phân \(\int\limits^{\pi}_0e^x.\sin x\text{d}x\) bằng
\(\dfrac{e^{\pi}-1}{2}\). \(\dfrac{e^{\pi}+1}{2}\). \(\dfrac{e^{\pi}+1}{4}\). \(\dfrac{e^{\pi}-1}{4}\). Hướng dẫn giải:\(\int\limits^{\pi}_0e^x.\sin x\text{d}x=\int\limits^{\pi}_0\left(e^x\right)'\sin x\text{d}x=e^x\sin x|_0^{\pi}-\int\limits^{\pi}_0e^x\left(\sin x\right)'\text{d}x=-\int\limits^{\pi}_0e^x\cos x\text{d}x\)
\(=\int\limits^{\pi}_0\left(e^x\right)'\left(-\cos x\right)\text{d}x=e^x\left(-\cos x\right)|_0^{\pi}-\int\limits^{\pi}_0e^x\left(-\cos x\right)'\text{d}x\)
\(=e^{\pi}+1-\int\limits^{\pi}_0e^x\sin x\text{d}x\)
Do đó \(2\int\limits^{\pi}_0e^x\sin x\text{d}x=e^{\pi}+1\Rightarrow\int\limits^{\pi}_0e^x\sin x\text{d}x=\dfrac{e^{\pi}+1}{2}\).
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