Suggested languages for you:

Americas

Europe

Q. 52

Expert-verified
Found in: Page 353

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Calculate the exact value of each definite integral in Exercises 47–52 by using properties of definite integrals and the formulas in Theorem 4.13.${\int }_{6}^{1}\left(3\left(1-2x{\right)}^{2}+4x\right)dx$

The exact value of definite integral is $-735$.

See the step by step solution

## Step 1. Given Information

We are given,

${\int }_{6}^{1}\left(3\left(1-2x{\right)}^{2}+4x\right)dx$

## Step 2. Finding the Integral

The definite integral is given by,

${\int }_{6}^{1}\left(3\left(1-2x{\right)}^{2}+4x\right)dx={\int }_{6}^{1}\left(3\left(1-4x+4{x}^{2}\right)+4x\right)dx\phantom{\rule{0ex}{0ex}}={\int }_{6}^{1}\left(3-12x+12{x}^{2}+4x\right)dx\phantom{\rule{0ex}{0ex}}=3{\int }_{6}^{1}1dx-8{\int }_{6}^{1}xdx+12{\int }_{6}^{1}{x}^{2}dx\phantom{\rule{0ex}{0ex}}=3\left[1-6\right]-8\left[\frac{1}{2}\left(1-36\right)\right]+12\left[\frac{1}{3}\left(1-216\right)\right]\phantom{\rule{0ex}{0ex}}=-15-4\left(-35\right)+4\left(-215\right)\phantom{\rule{0ex}{0ex}}=-15+140-860\phantom{\rule{0ex}{0ex}}=-735$