a.
Diện tích hình cần tìm là:
$S=\int\limits_a^{3a}\frac{4}{x}dx$
    $=4\ln x \left|\begin{array}{l}3a\\a\end{array}\right.$
    $=4\ln 3$

b,
Diện tích hình cần tìm là:
$S=\int\limits_{-2}^{4}(2x^2-4x-6)dx$
    $=\int\limits_{-2}^{-1}(2x^2-4x-6)dx-\int\limits_{-1}^3(2x^2-4x-6)dx+\int\limits_3^4(2x^2-4x-6)dx$
    $=\left(\frac{2x^3}{3}-2x^2-6x\right)\left|\begin{array}{l}-1\\-2\end{array}\right.-\left(\frac{2x^3}{3}-2x^2-6x\right)\left|\begin{array}{l}3\\-1\end{array}\right.+\left(\frac{2x^3}{3}-2x^2-6x\right)\left|\begin{array}{l}4\\3\end{array}\right.$
    $=\frac{14}{3}+\frac{64}{3}+\frac{14}{3}=\frac{92}{3}$

a.
Diện tích hình cần tìm là:
$S=\int\limits_1^e\frac{\ln x}{2\sqrt x}dx$
    $=\int\limits_1^e\ln xd(\sqrt x)$
    $=\sqrt x.\ln x \left|\begin{array}{l}e\\1\end{array}\right.-\int\limits_1^e\sqrt xd(\ln x)$
    $=\sqrt e-\int\limits_1^e\frac{1}{\sqrt x}dx$
    $=\sqrt e-2\sqrt x \left|\begin{array}{l}e\\1\end{array}\right.=2-\sqrt e$

b,
Diện tích hình cần tìm là:
$S=\int\limits_0^1x(x+1)^5dx$
    $=\int\limits_0^1(x^6+5x^5+10x^4+10x^3+5x^2+x)dx$
    $=\left(\frac{x^7}{7}+\frac{5x^6}{6}+2x^5+\frac{5x^4}{2}+\frac{5x^3}{3}+\frac{x^2}{2}\right)\left|\begin{array}{l}1\\0\end{array}\right.$
    $=\frac{107}{14}$

a,
Diện tích hình cần tìm là:
$S=\int\limits_{-2}^2|x(x+1)(x-2)|dx$
    $=-\int\limits_{-2}^{-1}x(x+1)(x-2)dx+\int\limits_{-1}^{0}x(x+1)(x-2)dx-\int\limits_{0}^{2}x(x+1)(x-2)dx$
    $=-\int\limits_{-2}^{-1}(x^3-x^2-2x)dx+\int\limits_{-1}^{0}(x^3-x^2-2x)dx-\int\limits_{0}^{2}(x^3-x^2-2x)dx$
    $=-\left(\frac{x^4}{4}-\frac{x^3}{3}-x^2\right)\left|\begin{array}{l}-1\\-2\end{array}\right.+\left(\frac{x^4}{4}-\frac{x^3}{3}-x^2\right)\left|\begin{array}{l}0\\-1\end{array}\right.-\left(\frac{x^4}{4}-\frac{x^3}{3}-x^2\right)\left|\begin{array}{l}2\\0\end{array}\right.$
    $=\frac{37}{12}+\frac{5}{12}+\frac{8}{3}=\frac{37}{6}$

b.
Diện tích hình cần tìm là:
$S=\int\limits_1^2\frac{1}{x(1+x^3)}dx$
Đặt: $x^3=t\Rightarrow dt=3x^2dx$
Đổi cận: $x=1\Rightarrow t=1$
                $x=2\Rightarrow t=8$
Ta có:
$S=\int\limits_1^2\frac{3x^2dx}{3x^3(1+x^3)}$
    $=\frac{1}{3}\int\limits_1^8\frac{dt}{t(1+t)}$
    $=\frac{1}{3}\int\limits_1^8(\frac{1}{t}-\frac{1}{t+1})dt$
    $=\frac{1}{3}\ln\frac{t}{t+1} \left|\begin{array}{l}8\\1\end{array}\right.=\frac{2}{3}\ln\frac{4}{3}$

a.
Diện tích hình cần tìm là:
$S=\int\limits_0^{\pi/2}\sin^2x\cos^3xdx$
    $=\int\limits_0^{\pi/2}\sin^2x(1-\sin^2x)d(\sin x)$
    $=\int\limits_0^{\pi/2}(\sin^2x-\sin^4x)d(\sin x)$
    $=\frac{\sin^3x}{3} \left|\begin{array}{l}\frac{\pi}{2}\\0\end{array}\right.-\frac{\sin^5x}{5} \left|\begin{array}{l}\frac{\pi}{2}\\0\end{array}\right.=\frac{2}{15}$

Bài 2:
b.
Ta có:
$\frac{x^2-59}{x+8}=\frac{x^2-64+5}{x+8}=x-8+\frac{5}{x+8}$
Suy ra:
$\frac{x^2-59}{x+8}\in\mathbb{Z} \Leftrightarrow \frac{5}{x+8}\in\mathbb{Z}$
                    $\Leftrightarrow \left[\begin{array}{l}x+8=1\\x+8=-1\\x+8=5\\x+8=-5\end{array}\right.$
                    $\Leftrightarrow \left[\begin{array}{l}x=-7\\x=-9\\x=-3\\x=-13\end{array}\right.$

Bài 2:
a.
Ta có:
$x^2-x+1=(x-\frac{1}{2})^2+\frac{3}{4}>0,\forall x.$
Để $\frac{7}{x^2-x+1}\in\mathbb{Z}$
thì $\left[\begin{array}{l}x^2-x+1=1\\x^2-x+1=7\end{array}\right.$
$\Leftrightarrow \left[ \begin{array}{l} x^2-x=0\\x^2-x-6=0 \end{array} \right.$
$\Leftrightarrow \left[\begin{array}{l}x=0\\x=1\\x=3\\x=-2\end{array}\right.$, thỏa mãn.

Bài 1:
Ta có:
$x+y+z=0\Rightarrow (x+y+z)^2=0$
                            $\Leftrightarrow x^2+y^2+z^2=-2(xy+yz+xz)$
Khi đó:
     $(x-y)^2+(y-z)^2+(z-x)^2$
$=2(x^2+y^2+z^2)-2(xy+yz+zx)$
$=3(x^2+y^2+z^2)$
Suy ra: $\frac{x^2+y^2+z^2}{(x-y)^2+(y-z)^2+(z-x)^2}=\frac{x^2+y^2+z^2}{3(z^2+y^2+z^2)}=\frac{1}{3}$