Expert answers to "Evaluate each of the following integrals. ∫02(x2+2x−3)3(4x+4)dx"

Brittney Lord

Answered question

2021-06-03

Evaluate each of the following integrals.

\int_{0}^{2} (x^{2} + 2 x - 3)^{3} (4 x + 4) d x

Answer & Explanation

Latisha Oneil

Skilled2021-06-04Added 100 answers

Step 1
To evaluate each of the following integrals.
Step 2
Given that
$\int_{0}^{2} (x^{2} + 2 x - 3)^{3} (4 x + 4) d x$
$= \int_{0}^{2} 4 (x^{2} + 2 x - 3)^{3} (x + 1) d x$
Let $x^{2} + 2 x - 3 = t$
Then $\frac{d t}{d x} = 2 x + 2$
$d t = 2 (x + 1) d x$
Also at $x = 0$ we have $t = - 3$
At $x = 2$ we have $t = 2^{2} + 4 - 3 = 8 - 3 = 5$
Hence we have
$\int_{0}^{2} (x^{2} + 2 x - 3)^{3} (4 x + 4) d x$
$= 2 \int_{- 3}^{5} t^{3} d t$
$= 2 [\frac{t^{4}}{4}]_{- 3}^{5} = \frac{1}{2} [5^{4} - (- 3)^{4}]$
$= \frac{1}{2} [625 - 81] = 272$
$∴ \int_{0}^{2} (x^{2} + 2 x - 3)^{3} (4 x + 4) d x = 272$

Jeffrey Jordon

Expert2021-11-19Added 2605 answers

Answer is given below (on video)

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Integral Calculus

Didn't find what you were looking for?