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Q. 52

Expert-verifiedFound in: Page 353

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.*

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

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

We are given,

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

The definite integral is given by,

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

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