Question

me ajudem pf n estou conseguindo resolver essa questao

Answer 1

Resposta:

$\int_0^2(4x^3-3x^2+2x)\;dx=12$

Explicação passo-a-passo:

$\int_0^2(4x^3-3x^2+2x)\;dx=\int_0^2(4x^3)\;dx+\int_0^2(-3x^2)\;dx+\int_0^2(2x)\;dx$

$\int_0^2(4x^3-3x^2+2x)\;dx=\left[4\cdot\frac{x^4}{4}\right]_0^2+\left[-3\cdot\frac{x^3}{3}\right]_0^2+\left[2\cdot\frac{x^2}{2}\right]_0^2$

$\int_0^2(4x^3-3x^2+2x)\;dx=\left[x^4\right]_0^2+\left[-x^3\right]_0^2+\left[x^2\right]_0^2$

$\int_0^2(4x^3-3x^2+2x)\;dx=(2^4-0^4)+\left[-2^3-(-0^3)\right]+(2^2-0^2)$

$\int_0^2(4x^3-3x^2+2x)\;dx=16-8+4$

$\int_0^2(4x^3-3x^2+2x)\;dx=12$