interdicoxd

2021-12-31

Evaluate the integral.
${\int }_{-8}^{4}\left(-2x+8\right)dx$

scomparve5j

Step 1
Given definite integral is:
${\int }_{-8}^{4}\left(-2x+8\right)dx$
We have to evaluate the above definite integral.
Step 2
Then we get,
${\int }_{-8}^{4}\left(-2x+8\right)dx$
$={\int }_{-8}^{4}\left(-2x\right)dx+{\int }_{-8}^{4}8dx$
$=-2{\int }_{-8}^{4}xdx+8{\int }_{-8}^{4}dx$
$=-2{\left[\frac{{x}^{2}}{2}\right]}_{-8}^{4}+8{\left[x\right]}_{-8}^{4}$
$=-2×\frac{1}{2}×\left[{4}^{2}-{\left(-8\right)}^{2}\right]+8\left[4-\left(-8\right)\right]$
=-(16-64)+8(4+8)
=-16+64+96
=144

kaluitagf

${\int }_{-8}^{4}-2x+8dx$
$\int -2x+8dx$
Use properties
$-\int 2xdx+\int 8dx$
Evaluate
$-{x}^{2}+8x$
Return the limits
$\left(-{x}^{2}+8x\right){\mid }_{-8}^{4}$
Calculate the expression
$-{4}^{2}+8\cdot 4-\left(-{\left(-8\right)}^{2}+8\cdot \left(-8\right)\right)$
144

Vasquez

Step 1
Given
$\int \left(-2x+8\right)dx$
Step 2
Solution:
$\int \left(-2x+8\right)dx=-{x}^{2}+8x$
Let's calculate a definite integral:
${\int }_{-8}^{4}\left(-2x+8\right)dx=\left(-{x}^{2}+8x\right){|}_{-8}^{4}$
F(4)=16
F(-8)=-128
I=16-(-128)=144
NSK