Question

19-42 evaluate the integral. 23. B2%7D%2B1%5Cright%29%20d%20x" id="TexFormula1" title="\int_{0}^{2}(2 x-3)\left(4 x^{2}+1\right) d x" alt="\int_{0}^{2}(2 x-3)\left(4 x^{2}+1\right) d x" align="absmiddle" class="latex-formula">

Answer 1

Expand the integrand:

$(2x - 3) (4x^2 + 1) = 8x^3 - 12x^2 + 2x - 3$

Then integrate using the power rule:

$\displaystyle \int (8x^3-12x^2+2x-3) \, dx = 2x^4 - 4x^3 + x^2 - 3x + C$

By the fundamental theorem of calculus, the definite integral has a value of

$\displaystyle \int_0^2 (2x-3)(4x^2+1) \, dx = \left(2x^4 - 4x^3 + x^2 - 3x\right)\bigg|_{x=2} - \left(2x^4 - 4x^3 + x^2 - 3x\right)\bigg|_{x=0}$

$\displaystyle \int_0^2 (2x-3)(4x^2+1) \, dx = 32 - 32 + 4 - 6 = \boxed{-2}$