Đáp án: $\displaystyle\int\limits^{10}_1f(t)dt=\dfrac{135}{4}$
Giải thích các bước giải:
Ta có:
$f(x^3+2x-2)=3x-1$
$\to (3x^2+2)f(x^3+2x-2)=(3x^2+2)(3x-1)$
$\to \displaystyle\int(3x^2+2)f(x^3+2x-2)dx=\displaystyle\int (3x^2+2)(3x-1)dx$
$\to \displaystyle\int f(x^3+2x-2)d(x^3+2x-2)=\displaystyle\int \:9x^3-3x^2+6x-2dx$
$\to \displaystyle\int f(x^3+2x-2)d(x^3+2x-2)=\dfrac{9x^4}{4}-x^3+3x^2-2x+C$
$\to \displaystyle\int\limits^2_1f(x^3+2x-2)d(x^3+2x-2)=\dfrac{135}{4}$
Đặt $t=x^3+2x-2\to t: 1\to 10$ vì $x: 1\to 2$
$\to \displaystyle\int\limits^{10}_1f(t)dt=\dfrac{135}{4}$
