# Can someone show me what to do with these Fund. Theorem Of Calc problems?

I have 3 problems dealing with the Fundamental Theorem of Calculus that I'm working on. The first one is relatively easy, but the second two are above my head. If anyone has the time to explain them, I would be thankful.

1: $\int_1^8 [3x^3 - \frac{5}{x} + 2 \sqrt[3]{x^2}$

My Answer:

$\int [ \frac{3*(8)^4}{4} - 5 ln(8) + \frac{2(8)^{3/2}}{3/2}] - \int [ \frac{3(1)^4}{4} - 5 ln(1) + \frac{2(1)^{5/2}}{5/2}$

2: $\int_0^{\pi/6} (sin(3x) - cos(3x))$

My answer:

I don't have one yet, I'm unsure of how to get the anti-derivative of a trig function with a variable larger than x.

3: $\int_{ln(1)}^{ln(2)} \frac{e^{2x}}{e^{2x} + 4}$

My answer:

I don't know where to begin with this one. Do I do u = $e^{2x} +4$ and du = $\frac{e^{2x}}{2}$?

EDIT

Updated answers:

1: $\int [ \frac{3*(8)^4}{4} - 5 ln(8) + \frac{2(8)^{5/3}}{5/3}] - \int [ \frac{3(1)^4}{4} - 5 ln(1) + \frac{2(1)^{5/3}}{5/3}$

2: $\int_0^{\pi/6} \frac{1}{3}cos(3(\frac{\pi}{6})) + \frac{1}{3}sin(3(\frac{\pi}{6}))$

3:

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## 1 Answer

Exercise 1:

The basic idea is right, your notation is weird though. Once you've found an antiderivative and are going to evaluate it at the endpoints of the integral, you don't write the integral sign anymore. It looks like this (for example):

$$\int_1^5 \frac{dx}{x} = \ln 5 - \ln 1 = \ln 5$$

Also, there's a mistake: $\sqrt[3]{x^2}$ is $x^{2/3}$, so what's its antiderivative?

Exercise 2:

Here the problem is finding the primitive of something like $\cos (3x)$. If that was $\cos x$, it would be easy: the primitive would just be $\sin x$, since $(\sin x)' = \cos x$. But there's a $3$ there, so let's try something involving a $3$. $3 \sin x$ is not going to work, because its derivative is $3 \cos x$. What about $\sin(3x)$? Its derivative is $3 \cos (3x)$, so we're getting there! Can you think of a function similar to $\sin(3x)$ such that its derivative is $\cos(3x)$?

Exercise 3:

Try a substitution $u = e^{2x}$. Then $du = 2e^{2x}\ dx = 2u\ dx$. Remember to change the bounds of integration accordingly. Since your substitution is $x \mapsto e^{2x}$, $\ln 2$ should be changed to $e^{2\ln 2}$ (this can be simplified) and similarly for $\ln 1$ which, by the way, equals $0$.

Clarification: The idea is that when you do a substitution, the bounds of the integral have to change. Let's say you have $\int_1^4 \frac{2x+1}{x^2+x}\ dx$. You can use a substitution $u = x^2+x$, $du = (2x+1)\ dx$. The integrand then becomes $\frac{du}{u}$. But it's not right to say that $\int_1^4 \frac{2x+1}{x^2+x}\ dx = \int_1^4 \frac{du}{u}$, because you haven't changed the limits of integration. Since your substitution is $u(x) = x^2+x$, your new limits will be $u(1)$ and $u(4)$, that is, $2$ and $20$. So $\int_1^4 \frac{2x+1}{x^2+x}\ dx = \int_2^{20} \frac{du}{u} = \ln u |_2^{20} = \ln 20 - \ln 2$.

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#1: $x^{2/3} -> x^{5/3}$. #2: Divide by 3? #3: I'm not sure I understand, so after I do u substitution, I should be plugging in $e^{2ln(2)}$ and $2 ln(1)$? – StrugglingWithMath Dec 10 '12 at 21:14
1: Remember to divide, too: $\int x^{2/3}\ dx = \frac{x^{5/3}}{5/3} + C$. 2: Yes, that's right. As always, you can check: $(\frac{1}{3}\sin (3x))' = \frac13 3 \cos (3x) = \cos(3x)$. 3: I'll edit more information. – Javier Dec 10 '12 at 21:19
Ah, thank you for clarifying. I didn't realize I had to plug the values into u. So if u = $e^{2x}$, I plug $ln(1)$ and $ln(2)$ into it, so my new bounds are: a = 1, b = 2? If so, then I just struggle on what the equation is supposed to look like. – StrugglingWithMath Dec 10 '12 at 21:39
Almost. $e^{2\ln 2} = e^{\ln (2^2)} = 2^2 = 4$. But otherwise that's right, yes. – Javier Dec 10 '12 at 21:46
Thank you for your patience and kindness in helping me. Do you mind showing me what the equation should be to plug the new bounds into? – StrugglingWithMath Dec 10 '12 at 21:48