`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).
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nees101 | Student, Graduate | (Level 2) Adjunct Educator

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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`

So, the answer is -2.

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scisser | (Level 3) Honors

Posted on

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`

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