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

### Textbook Question

Chapter 5, 5.4 - Problem 25 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

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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

scisser | (Level 3) Honors

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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