Warmup: Definite integral properties (no graph)

Let's imagine what $f(x)$‍ may look like and think of these definite integrals as net areas. We can add the net areas of smaller intervals to get the net area of the entire interval.

We know that the net area from $x=-5$‍ to $x=5$‍ is $10$‍ since ${\displaystyle {\int }_{-5}^{5}f(x)dx=10}$‍.

What does the net area have to be from $x=-1$‍ to $x=-3?$‍

We can add the net areas of smaller intervals to get the net area of a larger interval.

Multiplying $f(x)$‍ by $7$‍ makes the function and net area $7$‍ times as large in the given interval.

So we can take a constant multiplier of $f(x)$‍ out in front of the integral.

We can reverse the order of the bounds on an integral, and this makes the entire integral negative.

We can write this as two separate integrals added together.

Finding ${\displaystyle {\int }_3^{5}f(x)dx:}$‍
It was given that ${\displaystyle {\int }_3^{5}f(x)dx}=9$‍

Finding ${\displaystyle {\int }_3^{5}4dx:}$‍
Let's use the graph of $y=4$‍ and look at the net area from $x=3$‍ to $x=5.$‍

The net area is a rectangle.

Combine the areas:

Let's imagine what $g(x)$‍ may look like from $x=-8$‍ to $x=0$‍ and think of these definite integrals as net areas. We can add the net areas of smaller intervals to get the net area of a larger interval.

We know that the net area from $x=-8$‍ to $x=0$‍ is $5$‍ since ${\displaystyle {\int }_{-8}^{0}g(x)dx=5}$‍.

What does the net area have to be from $x=-2$‍ to $x=0?$‍

Let's imagine what $g(x)$‍ may look like from $x=-2$‍ to $x=8$‍ and think of these definite integrals as net areas. We can add the net areas of smaller intervals to get the net area of a larger interval.

We know that the net area from $x=-2$‍ to $x=8$‍ is $-5$‍ since ${\displaystyle {\int }_{-2}^{8}g(x)dx=-5}$‍.

Here are the net areas we've found so far.

Let's take the $10$‍ out to the front of the integral, switch the bounds, and break up the interval.

We can write this as two separate integrals added together.

Finding ${\displaystyle {\int }_0^{8}g(x)dx:}$‍
We already found that ${\displaystyle {\int }_0^{8}g(x)dx}=1$‍

Finding ${\displaystyle {\int }_0^{8}xdx:}$‍
Let's use the graph of $y=x$‍ and look at the net area from $x=0$‍ to $x=8.$‍

The net area is a triangle.

Combine the areas: