Q. 52 - Calculus

Question

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.

${\int }_6^1\left(3(1-2x{)}^2+4x\right)dx$

Step-by-Step Solution

Answer

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

Step 1. Given Information

We are given, 

${\int }_6^1\left(3(1-2x{)}^2+4x\right)dx$

Step 2. Finding the Integral

The definite integral is given by,

 $$\begin{gathered} {\int }_6^1\left(3(1-2x{)}^2+4x\right)dx={\int }_6^1\left(3\left(1-4x+4x^2\right)+4x\right)dx \\ ={\int }_6^1\left(3-12x+12x^2+4x\right)dx \\ =3{\int }_6^{1}1dx-8{\int }_6^{1}xdx+12{\int }_6^1{x}^{2}dx \\ =3[1-6]-8\left[\frac{1}{2}(1-36)\right][phone]\mathrm{}\left[\frac{\mathrm{}}{\mathrm{}}\mathrm{}\mathrm{})\right] \\ =-15-4(-35)+4(-215) \\ =-15+140-860 \\ =-735 \end{gathered}$$