How to Integrate g(x)=∫ƒ(t) dt from -5 to x | Homework Help

[romeIAM]

Homework Statement

http://s23.postimg.org/wsj9e91wb/IMG_1334.jpg[/B]
photo of the problem

g(x)=∫ƒ(t) dt from -5 to x

ƒ(t) = (0 if x < -5
5 if -5≤x<-1
-3 if -1≤≤x<3
0 if x≥3)

(a) g(-8) = 0
(b) g(-4) = 5
(c) g(0) = ?
(d) g(4) = ?

Homework Equations

∫ƒ(x) from a to b = (f'(b)-f'(a))
fundamental theorem of calculus

The Attempt at a Solution

[/B]
I was able to get that g(-8) = 0 because plugging -8 into the upper limit and meant x was less than -1 and gave me with f(t) equaling 0 and the anti derivative of 0 is 0. Now i got g(-4) = 5 because the antiderivative of 5 is 5x and plugging in -4 → x and -5→x got me 5. However when i do the same for c and i get -15 by plugging in 0 to x making the f(t) = -3 . The last 0 i thought the anti derivative of 0 is 0 but it doesn't take either of those answers. I need guidence.

 

[Mark44]

Homework Statement

http://s23.postimg.org/wsj9e91wb/IMG_1334.jpg[/B]
photo of the problem

g(x)=∫ƒ(t) dt from -5 to x

ƒ(t) = (0 if x < -5
5 if -5≤x<-1
-3 if -1≤≤x<3
0 if x≥3)

(a) g(-8) = 0
(b) g(-4) = 5
(c) g(0) = ?
(d) g(4) = ?

Homework Equations

∫ƒ(x) from a to b = (f'(b)-f'(a))
fundamental theorem of calculus

This is NOT what the FTC says!

The Attempt at a Solution

[/B]
I was able to get that g(-8) = 0 because plugging -8 into the upper limit and meant x was less than -1 and gave me with f(t) equaling 0 and the anti derivative of 0 is 0. Now i got g(-4) = 5 because the antiderivative of 5 is 5x and plugging in -4 → x and -5→x got me 5. However when i do the same for c and i get -15 by plugging in 0 to x making the f(t) = -3 . The last 0 i thought the anti derivative of 0 is 0 but it doesn't take either of those answers. I need guidence.

Much better! Thank you.

Sketch the graph, if you haven't already done so. If you have the graph, this is a very simple problem.
For c) g(0) = $\int_{-5}^{-1} f(t)dt + \int_{-1}^0 f(t)dt$
The first integral is positive, since the graph of f is above the horizontal axis. The second integral is negative, because the graph is below the hor. axis.

For d), the approach is similar.