Question

Please help me with my calculus homework, question 4 only***

Answer 1

The given integral is

`\int ^3_0(x^3-6x)dx`

Recall the fundamental theorem of calculus,

`\int ^b_af(x)dx=F(b)-F(a)`

The given intergral is

`\int ^b_af(x)dx=\int ^3_0(x^3-6x)dx`
`F(x)=\int f(x)dx=\int (x^3-6x)dx`

Integrating with respect to x, we get

`F(x)=(x^4)/(4)-(6x^2)/(2)`

The antiderivative of f is

`F(x)=(x^4)/(4)-3x^2`

Replace x=3 in F(x) to compute F(3).

`F(3)=((3)^4)/(4)-3(3)^2=20.25-27=6.75`

Replace x=0 in F(x) to compute F(0).

`F(0)=((0)^4)/(4)-3(0)^2=0`

The given integral is

`\int ^3_0(x^3-6x)dx=F(3)-F(0)`

Substitute F(3)=6.75 and F(0)=0, we get

`\int ^3_0(x^3-6x)dx=6.75-0`

Hence the answer is

`\int ^3_0(x^3-6x)dx=6.75=6(3)/(4)`