Find the area under the graph of the function $f(x)=4x-7$ between $x=-2$ and $x=0$ Section 5.4

Can somebody verify this solution for me? Thanks!

Find the area under the graph of the function $f(x)=4x+7$ between $x=-2$ and $x=0$



Well, the area under the graph of $f(x)=4x+7$ between $x=-2$ and $x=0$ is exactly equal to:

$\int_{-2}^0 4x+7dx$

$= [\frac{4}{2}x^2+7x] |_{-2}^0$

$= [2x^2+7x] |_{-2}^0$

$=(2(0)^2+7(0))-(2(-2)^2+7(-2))$

$=0-(2(4)-14)$

$=6$
You can easly check it your self in this case. Just calulate the area of triangles:
You don't need integral since it's a linear equation. You can just find the area under the triangle as @Aqua mentioned. Base is -2, and height is 7 since $f(x)$ is 0. Then multiply and find the area.
$$
\frac{-2\times7}{2}=-7
$$