MHB Integrating a piecewise function?

[Emma_011]

I have to find:

g(1)=
and
g(5)=

I have drawn the graph and I am a little unsure where to go from there. I know area is involved somehow but not entirely sure what to do. Any help is appreciated

 

[I like Serena]

View attachment 10849

I have to find:

g(1)=
and
g(5)=

I have drawn the graph and I am a little unsure where to go from there. I know area is involved somehow but not entirely sure what to do. Any help is appreciated

Welcome to MHB Emma! ;)

We have a rectangular area between x=-5 and x=0 in the graph that is above the x-axis.
What is its area?
We have another rectangular area between x=0 and x=1 that is below the x-axis.
What is its area?

The value of g(1) is the area above the x-axis (starting from x=-5) minus the area below the x-axis (up to x=1). Can you find it?

 

[Emma_011]

Based on what is given, the area would be 20 between x=-5 and x=0 and the area between x=0 and x=1 would be 5.

So it would be 20-5=15

 

[I like Serena]

Based on what is given, the area would be 20 between x=-5 and x=0 and the area between x=0 and x=1 would be 5.

So it would be 20-5=15

Yep. (Nod)

For g(5) the negative area runs from x=0 to x=4.
After that we have zero (up to x=5), so there is no area there to add or subtract.
Can we find g(5) then?

 

[Emma_011]

Since there is no area to add or subtract would it just be 0?

 

[I like Serena]

Since there is no area to add or subtract would it just be 0?

No, only the part to the right of x=4 is zero.
We still have the positive rectangle between x=-5 and x=0, an the negative rectangle between x=0 and x=4.
So g(5) is the difference between the areas of those two rectangles.

 

[HOI]

Frankly, I wouldn't have thought of this as "area" at all but just use the basic rule of itegration: $\int_a^b f(x)dx= \int_a^c f(x)dx+ \int_c^b f(x)dx$.

$g(x)= \int_{-5}^x f(x)dx$ is:
if $x\le -5$, $g(x)= \int_{-\infty}^x 0 dx= 0$
if $-5\le x\le 0$ $g(x)= \int_{-5}^x 4dx= \left[4x\right]_{-5}^x= 0- 4x= -4x$
if $0\le x\le 4$ $g(x)= \int_{-5}^0 4dx+ \int_0^x -5 dx= -4(-5)+ \left[-5x\right]_0^x= 20+ -5x- 0= 20- 5x$
if $x\ge 5$ $g(x)= \int_{-5}^4 f(x)dx+ \int_4^5 0 dx= 20- 5(4)= 0$.