Question

[Ngọc Ánh Nguyễn Thị]

1/ I=\(\int_{-2}^2\left|x^2-1\right|dx\)

2/ I= \(\int_1^e\sqrt{x}.lnxdx\)

3/ I= \(\int_0^{\dfrac{\pi}{2}}\left(e^{sinx}+cosx\right)cosxdx\)

4/ I= \(\int_0^{\dfrac{pi}{2}}\dfrac{sin2x}{\sqrt{cos^2x+4sin^2x}}dx\)

5/ I= \(\int_0^{\dfrac{\pi}{4}}\sqrt{2}cos\sqrt{x}dx\)

6/ I= \(\int_1^{\sqrt{e}}\dfrac{1}{x\sqrt{1-ln^2x}}dx\)

7/ I= \(\int_{-\dfrac{\pi}{4}}^{\dfrac{\pi}{4}}\dfrac{sin^6x+cos^6x}{6^x+1}dx\)

Answer 1

[Hoàng Tử Hà]

Nhìn đề dữ dội y hệt cr của tui z :( Để làm từ từ 

Lập bảng xét dấu cho \(\left|x^2-1\right|\) trên đoạn \(\left[-2;2\right]\)

x	-2	-1	1	2
\(x^2-1\)		0	0

\(\left(-2;-1\right):+\)

\(\left(-1;1\right):-\)

\(\left(1;2\right):+\)

\(\Rightarrow I=\int\limits^{-1}_{-2}\left|x^2-1\right|dx+\int\limits^1_{-1}\left|x^2-1\right|dx+\int\limits^2_1\left|x^2-1\right|dx\)

\(=\int\limits^{-1}_{-2}\left(x^2-1\right)dx-\int\limits^1_{-1}\left(x^2-1\right)dx+\int\limits^2_1\left(x^2-1\right)dx\)

\(=\left(\dfrac{x^3}{3}-x\right)|^{-1}_{-2}-\left(\dfrac{x^3}{3}-x\right)|^1_{-1}+\left(\dfrac{x^3}{3}-x\right)|^2_1\)

Bạn tự thay cận vô tính nhé :), hiện mình ko cầm theo máy tính 

Answer 2

[Hoàng Tử Hà]

2/ \(I=\int\limits^e_1x^{\dfrac{1}{2}}.lnx.dx\)

\(\left\{{}\begin{matrix}u=lnx\\dv=x^{\dfrac{1}{2}}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{2}{3}.x^{\dfrac{3}{2}}\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}\int\limits^e_1x^{\dfrac{1}{2}}.dx\)

\(=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}.\dfrac{2}{3}.x^{\dfrac{3}{2}}|^e_1=...\)