# Homework Help: Integration Questions

1. Jan 25, 2005

Hello all

Here are a few problems I encountered:

Find the area bounded by the parabola $$y = \frac{1}{2}x^2 + 1$$ and the straight line $$y = 3 + x$$ Ok, so I first set the two equations equal to each other:

$$3 + x = \frac{1}{2}x^2 + 1$$ After solving for x, do I just use those two values for the upper and lower limits, and evaluate:

$$\int^b_a \frac{1}{2}x^2 + x - 2$$?

Also how would you evalutate the following integrals:

$$\int^b_a (x+1)^a dx$$
$$\int^b_a sin \alpha x {} dx$$
$$\int^b_a cos \alpha x {} dx$$

For the first one would I use a geometric progression? I know that for $$\int^b_a x^a dx$$ you divide up the interval using the following points:
$$a, aq, aq^2, . . . , aq^n^-1, aq^n = b$$ So that means for $$a+1$$ we have $$a+1, (a+1)q, (a+1)q^2, . . . (a+1)q^n-1, (a+1)q^n$$?

For the other integrals, do I just make use of the identity $$2\sin u sin v = \cos(u-v) - \cos(u+v)$$?

Thanks

Last edited: Jan 25, 2005
2. Jan 25, 2005

### dextercioby

1.Why did you set equal those 2 equations involving 2 parabolas??Shouln't have it been a parabola & a line??
2.The first integral is realized through an elementary substitution...Do you see it?
3.The second one is accomplished through a substitution as well.
4.The third one is similar to the second one... Meaning you need to do the same substitution...
Daniel.

3. Jan 25, 2005

yes but I am not supposed to use substitutions. I have to actually use the limiting process! This is the beginning of the integrals chapter.

4. Jan 25, 2005

### dextercioby

Okay,u should have said that from the beginning...What about the first exercise??Why 2 parabolas??

Daniel.

P.S.U edited your message... :tongue2: Solve for "x" and then determine what curve is on "top" of the other,compute the 2 areas and then subtract them (if they're both positive/negative) or add them if one of them is negative.

5. Jan 25, 2005

Thanks. Any help or hints for the last few problems are appreciated.

Also if we have to evaluate $$\int^1_0 (1-x)^n dx$$ where n is an integer do I just expand the bracket? Would it be $$1 - nx + x^n +...$$?

Thanks

6. Jan 25, 2005

### dextercioby

Without substitution??Okay,the u'd have to use the binomial formula:
$$(1-x)^{n}=\sum_{k=0}^{n} C_{n}^{k} (-1)^{n-k} x^{n-k}$$

Daniel.

7. Jan 25, 2005

ok so we know $$k = 0, n = 1$$. So it would be $$C^0_1 (-1)^1 x^1 = -x$$ So is is $$\int^1_0 -x dx = \frac {-1}{2}(b^2 - a^2) = \frac {-1}{2}$$?

Thanks

Last edited: Jan 25, 2005
8. Jan 25, 2005

I think it's $$\frac {-1}{2}$$ because I just plugged in the values.

Is this right?

9. Jan 25, 2005

### dextercioby

$$\int_{0}^{1} -x \ dx$$

is equal to $$-\frac{1}{2}$$
??If so,then the answer is YES.

Daniel.

10. Jan 25, 2005

thank you. is that the answer to $$\int^1_0 (1-x)^n$$ after simplifying?

11. Jan 26, 2005

### dextercioby

$$\int_{0}^{1} (1-x)^{n} dx=\frac{1}{n+1}$$

Daniel.

12. Jan 26, 2005

how did you get this? Did you substitute the values for k and n?

Thanks

13. Jan 26, 2005

### dextercioby

Daniel.

14. Jan 26, 2005

I used the binomial formula and got $$_-x$$

15. Jan 26, 2005

### dextercioby

Daniel.

16. Jan 26, 2005

Because as you said using the binomial forumula $$(1-x)^{n}=\sum_{k=0}^{n} C_{n}^{k} (-1)^{n-k} x^{n-k}$$ shouldn't I just substitute $$k = 0, n = 1$$ and this equals $$1(-1^{1-0})(x^{1-0}) = - x$$

17. Jan 26, 2005

### dextercioby

No,"k" is a summation index.And "n" is natural,arbitrary and finite...

Daniel.

18. Jan 26, 2005

Then how would you use the binomial formula?

Thanks

19. Jan 26, 2005

### dextercioby

By writing explitely each term from the sum and then applying the fact that
$$\int_{a}^{b} [f(x)+g(x)] \ dx=\int_{a}^{b} f(x) \ dx +\int_{a}^{b} g(x) \ dx$$

The property of addition for the definite integrals...

Daniel.