# `int_0^2 (2x - 3)(4x^2 + 1) dx` Evaluate the integral

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### 2 Answers

We have to evaluate the integral:

`\int_{0}^{2}(2x-3)(4x^2+1)dx=\int_{0}^{2}(8x^3-12x^2+2x-3)dx`

`=[8(x^4/4)-12(x^3/3)+2(x^2/2)-3x]_{0}^{2}`

`=[2x^4-4x^3+x^2-3x]_{0}^{2}`

`=[2(2^4)-4(2^3)+(2^2)-3(2)` `=[2(2^4)-4(2^3)+2(2^2)-3(2)]-0`

`=2(16)-4(8)+4-6` `=2(16)-4(8)-2(4)-6`

`=32-32+4-6` `=32-32-8-6`

`=-2` `=-14`

So, the answer is -2.

Multiply the terms out

`int_0^2(8x^3-12x^2+2x-3)`

Find the antiderivative of each term: `int (a^n=(a^(n+1))/(n+1))`

`=2x^4-4x^3+x^2-3x|_0^2`

Plug in the upper limit and subtract away the lower limit

`[((2)(2)^4-4(2)^3+(2)^2-3(2)]-[(2(0)^4-4(0)^3+(0)^2-3(0)]`

`=-2`