Tich phan

Question

$I=\int\limits_{0}^{1}(x^{2}e^{x^{3}}+\frac{\sqrt[4]{x}}{1+\sqrt{x}})dx$

GreenmjlkTea FeelingTea · Answer 1 · 6/14/2013 9:13:48 PM

Bước 1. Tính

$\int x^2e^{x^3}dx=\frac{1}{3}\int d(x^3).e^{x^3}=\frac{1}{3}.e^{x^3}$ (PP đổi biến tích phân)

$\Rightarrow \int\limits_{0}^{1}x^2e^{x^3}dx=\frac{1}{3}(e^1-e^0)=\frac{1}{3}(e-1)$

Bước 2 . Tính

$\int\limits_{0}^{1} \frac{\sqrt[4]{x}}{1+\sqrt x}dx$ bằng phương pháp hữu tỉ hóa

Đặt

$u=\sqrt[4]{x}=x^\frac{1}{4}\in[0,1]\Rightarrow du=\frac{1}{4}x^{-\frac{3}{4}}dx$

$\Rightarrow dx=4du.x^\frac{3}{4}=4du.u^3$

$\int\limits_{0}^{1}\frac{\sqrt[4]{x}}{1+\sqrt x}dx=\int\limits_{0}^{1}\frac{u}{1+u^2}.4u^3.du=4\int\limits_{0}^{1}\frac{u^4}{1+u^2}du$

$=4\int\limits_{0}^{1}(\frac{u^4-1}{u^2+1}+\frac{1}{u^2+1})du$

$=4\int\limits_{0}^{1}(u^2-1)du+4\int\limits_{0}^{1}\frac{1}{u^2+1}du$

Do

$\int(u^2-1)du=\frac{1}{3}u^3-u , \int\frac{1}{u^2+1}=tan^{-1}u$ nên

$=4(\frac{1}{3}-1)+4.(tan^{-1}1-tan^{-1}0)$

$=4.-\frac{2}{3}+4.\frac{\pi}{4}$

$=\pi-\frac{8}{3}$

Bước 3. Kết luận

$\int\limits_{0}^{1}(x^2e^{x^3}+\frac{\sqrt[4]{x}}{1+\sqrt x})dx=\frac{1}{3}(e-1)+\pi-\frac{8}{3}=\frac{e}{3}+\pi-3$

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