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.

${\int }_0^4\left((2x-3{)}^2+5\right)dx$

Short answer

The exact value of definite integral is $\frac{136}{3}$.

Step-by-step solution

Step 1. Given Information  

We are given,

${\int }_0^4\left((2x-3{)}^2+5\right)dx$

Step 2. Finding the Integral  

The definite integral is given by,

$$\begin{gathered} {\int }_0^4\left((2x-3{)}^2+5\right)dx={\int }_0^4\left(4x^2-12x+9+5\right)dx \\ ={\int }_0^4\left(4x^2-12x+14\right)dx \\ =4{\int }_0^4{x}^{2}dx-12{\int }_0^{4}xdx+{\int }_0^{4}14dx \\ =4\left(\frac{1}{3}\left(4^3\right)\right)-12\left(\frac{1}{2}\right)\left(4^2\right)+14\left(4\right) \\ =4\left(\frac{64}{3}\right)-6\left(16\right)+56 \\ =\frac{256}{3}-96+56 \\ =\frac{256-288+168}{3} \\ =\frac{136}{3} \end{gathered}$$