How to Evaluate Integrals Using the Definite Integral Properties

[courtrigrad]

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

 

[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.

 

[courtrigrad]

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.

 

[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... 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.

 

[courtrigrad]

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

 

[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.

 

[courtrigrad]

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

 

[courtrigrad]

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

Is this right?

 

[dextercioby]

Are u asking whether
\int_{0}^{1} -x \ dx

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

Daniel.

 

[courtrigrad]

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

 

[dextercioby]

No,the answer is:

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

Daniel.

 

[courtrigrad]

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

Thanks

 

[dextercioby]

Nope,i made a substitution.

Daniel.

 

[courtrigrad]

I used the binomial formula and got _-x

 

[dextercioby]

What?Please explain your result...

Daniel.

 

[courtrigrad]

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

Thanks for your help

 

[dextercioby]

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

Daniel.

 

[courtrigrad]

Then how would you use the binomial formula?

Thanks

 

[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.