tolong ya akak. no 21 ya kk. beserta dengan penjelasan ya akak.

[tex]\displaystyle y_1=y_2\\x^2-2x=3x\\x^2-5x=0\\x(x-5)=0\\x=0\vee x=5\\\\A=\int\limits^{x_2}_{x_1}y_2-y_1\,dx\\A=\int\limits^{5}_{0}3x-x^2+2x\,dx\\A=\int\limits^{5}_{0}5x-x^2\,dx\\A=\left\frac52x^2-\frac13x^3\right|^{5}_{0}\\A=\frac52(5^2-0^2)-\frac13(5^3-0^3)\\A=\frac52(25-0)-\frac13(125-0)\\A=\frac{125}2-\frac{125}3\\A=\frac{125}6\\\boxed{\boxed{A=20\frac{5}6\,\text{sL}}}[/tex]

20) Lim (Tan ax)/(sin bx) = a/b
.......x=>0
Lim (Tan x)/(sin 2x) = 1/2
x=> 0

21) luas yg hanya dibatasi 2 kurva dengan y1 - y2 = ax^2 + bx + c adalah
L = (D√D)/(6a^2) dengan D = b^2 - 4ac

y1 - y2 = (x^2 - 2x) - 3x = x^2 - 5x
D = b^2 - 4ac = (-5)^2 - 4(1)(0) = 25
Luas = (D√D)/(6a^2) = (25√25)/(6(1)^2) = (25 . 5)/6 = 125/6 = 20 5/6

22) f(x) = 2x^(2/3) - 6x.x^1/2 = 2x^2/3 - 6x^3/2
f'(x) = 2(2/3)x^(2/3 - 1) - 6(3/2)x^(3/2 - 1)
= 4/3 x^(-1/3) - 9x^1/2
f'(1) = 4/3 (1)^-1/3 - 9(1)^1/2 = 4/3 - 9 = -23/3