Question

\(I= \int_{-2}^2\left(\sqrt{20-x^2}-x^2\right)dx\)

Answer 1

Áp dụng nguyên hàm cơ bản: \(\int\sqrt{a^2-x^2}dx=\dfrac{a\sqrt{a^2-x^2}}{2}+\dfrac{a^2}{2}arcsin\dfrac{x}{2}+C\)

\(I=\left(\dfrac{x\sqrt{20-x^2}}{2}+10arcsin\dfrac{x}{2\sqrt{5}}\right)|^2_{-2}-\dfrac{1}{3}x^3|^2_{-2}=...\)

Answer 2

\(I=\int\sqrt{20-x^2}dx-\int x^2dx\)

Xet \(I_1=\int\sqrt{20-x^2}dx\)

\(x=\sqrt{20}\sin t\left(-\dfrac{\pi}{2}\le t\le\dfrac{\pi}{2}\right)\Rightarrow dx=\sqrt{20}\cos tdt\)

\(\Rightarrow I_1=\int\sqrt{20\cos^2t}.\sqrt{20}\cos tdt=20\int\cos^2t.dt=10\int dt+10\int\cos2t.dt=10t+5\sin2t+C\)

\(\Rightarrow I=10arc\sin\left(\dfrac{x}{\sqrt{20}}\right)+5\sin\left[2.arc\sin\left(\dfrac{x}{\sqrt{20}}\right)\right]-\dfrac{1}{3}x^3+C\)

P/s: Bạn tự thay cận vô ạ