# Homework Help: Some antiderivative questions

1. May 5, 2007

### t_n_p

1. The problem statement, all variables and given/known data
Q1.
http://img523.imageshack.us/img523/5933/17857219ur6.jpg [Broken]

Q2.
http://img152.imageshack.us/img152/8133/32665282he1.jpg [Broken]

3. The attempt at a solution

I really don't know how to go about either of the two.
In the second one I know that g(0)=100, g'(0)=-10 and g'(15)=0, but have no idea how you would be able to use these go get g(15)

Last edited by a moderator: May 2, 2017
2. May 5, 2007

### gunstar-red

Q1 Simplify the integral using all those basic rules. The integral of a constant times a function is equal to a constant times the integral of the function. And the sum of the integrals is equal to the integral of the sum.
$$\int_{-1}^{1}(3f(x)-g(x))dx = 3\int_{-1}^{1}f(x) dx - \int_{-1}^{1} g(x) dx = 6-\int_{-1}^{1} g(x)\\ 6-\int_{-1}^{1} g(x) = 12\\ \int_{-1}^{1} g(x) = -6\\$$

Last edited: May 5, 2007
3. May 5, 2007

### t_n_p

whoops! I posted the wrong question. Thanks anyway for the help!

Here's the one I was supposed to post..

http://img329.imageshack.us/img329/6145/70949891jf2.jpg [Broken]

Last edited by a moderator: May 2, 2017
4. May 5, 2007

### gunstar-red

$$let  u=x^2,\frac{du}{dx}=2x \\  xdx =\frac{1}{2}du \\ when x =0, u=0\\ when x =2, u=4\\  \int_0^2 xg(x^2)dx=\int_0^4 \frac{1}{2} g(u) du = \frac{1}{2}*16=8$$

5. May 5, 2007

### t_n_p

Thanks alot of sorting that one out!

Now for that dreaded antidiv. graph..:grumpy:

6. May 5, 2007

### gunstar-red

I'll take a guess and say that the answer is 25, the curved section of the graph is basically a distraction. You can find the total change between 0 and 15 by finding the area under the derivative graph for this interval. It's equal to -75. As g(0) = 100 it follows that g(15) = 100 -75 =25.

7. May 5, 2007

### t_n_p

hmmm, I'm really clueless, but the theory behind your answer seems like the most logical/correct understanding I have heard so far.

8. May 5, 2007

### HallsofIvy

As gunstar-red said, the "looping" part at the end is irrelevant since you are only asked to go up to x= 15. The graph before that is a straight line from (0, -10) to (15, 0). Although it is harder than what he suggested, you can fairly easily write out the equation for g'(x) and integrate to find g(x).

9. May 5, 2007

### gunstar-red

I still get 25 doing it your way HallsofIvy.
g'(x)=2/3x-10
g(x)=x^3/3-10x+c
g(0)=100
.:100=c
g(x)=x^3/3-10x+100
g(15)=75-150+100=25

I think this is because they are in essence the same process.

g'(x)=f(x)
g(x)=F(x)+c when x = 0, g(0)=100
100=F(0)+c
c=100-F(0)
so g(x)=F(x) - F(0) +100
and g(15) =F(15) - F(0) +100
$$g(15) = \int_0^{15} g'(x) dx +100 = -75 +100 = 25$$

10. May 5, 2007

### HallsofIvy

Yes, of course, you get the same thing! And, yes, they are basically the same.

11. May 5, 2007

### t_n_p

HallsofIvy, your method makes alot more sense to me solely because theres more math and less theory!