Evaluate the integral. ?

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Answer 1 · 2017-12-14T01:29:27

#int_0^2(6x-9)(4x^2+7)dx=-42#

Multiply out #(6x-9)(4x^2+7)#

#=(24x^3+42x-36x^2-63)=24x^3-36x^2+42x-63#

The previous step makes the integration much simpler and clear and now are left with the integral:

#int_0^2 24x^3-36x^2+42x-63# #dx#

We will integrate each term (expect the last term) using the following rule of integration.

#intx^ndx=(x^(n+1))/(n+1); n!=1#

For the last term:

Recall #intkdx=kx+C#

Thus integrating,

#=(24x^4)/4-(36x^3)/3+(42x^2)/2-63x+"C"#

#=6x^4-12x^3+21x^2-63x+"C"#

Now evaluating the definite integral:

#[6x^4-12x^3+21x^2-63x+"C"]_0^2#

#=[6(2)^4-12(2)^3+21(2)^2-63(2)+"C"]-[6(0)^4-12(0)^3+21(0)^2-63(0)+"C"]#

#=-42-0#

#=-42#