1. ## Calculus Integral Problems

Hello,

Firstly I would like to say hello, and I'm new to these forums but I think I will use this forum very frequently

At the moment these two problems are difficult for me to solve.

First problem is:
(d/dx) ∫ (0 to tanx) e^(t^2)
Image of the problem http://i.imgur.com/D6Qh9.jpg

Second problem is:
the area of ∫ (-1 to 1) (x-3)/(x^2+1)
Image of the problem http://i.imgur.com/q8EKU.jpg

I think the answer to the first problem is e^[(tanx)^2] * (secx)^2, but unsure if that's actually correct.

I believe the answer to the second problem Is [(1/2)(x^2)-(3x)] * arctanx? Therefore 35pi/8 + 7pi/8 = 21pi/4, but I get the feeling that this is wrong.

Any help would be much appreciated. Thank you

2. ## Re: Calculus Integral Problems

Hi, dimelychime. Welcome to the forum!

The first solution looks correct (this is the Fundamental Theorem of Calculus along with the chain rule).

I got $-\frac{3\pi}{2}$ for the second problem and checked this a couple times. Note: I used the substitution $x=\tan\theta$ to find an antiderivative.

I didn't see an attachment with work for the second problem so I don't know where our differences may be.

Let me know if you try it again. If there are still differences, I will write up more details in a later post. Good luck!

3. ## Re: Calculus Integral Problems

Hello, thanks for the reply. Yes I got -3pi/2 on my second try [last step looked (-3)(pi/4)- (-3)(-pi/4)]. I'm now unsure when to apply absolute area rule (where you would have to consider everything positive) or just integrals (hope that makes sense).

Also, I'm pretty confident that this is the right answer to the following problem...
∫ {0,2} (2^x)+6 dx

[e^ln(2)x]+6 dx

>substitute xln(2) for u
dx=1/ln2 du

(1/ln2) ∫ {0,2} (e^u)+6 du
(1/ln2) ∫ {0,2} (e^(xln2))+6 du
(1/ln2)(2^x)+6x

then plug in 2 and 0.

4/ln2 - 1/ln2 = 3/ln2 <I think this is wrong
tried it again and I got 15/ln2

So is this right or is it unimaginably wrong?

Thank you

4. ## Re: Calculus Integral Problems

To answer your first question about absolute area, I would say that it should be stated in the directions of the exercise. If I see a problem that simply says evaluate some definite integral, then I don't take absolute value. However, if the problem says something like total area then I do.

For your second question, you're definitely on the right track. Rewriting $2^{x}=e^{x\ln 2}$ is a nice idea. You make the substitution $u=x\ln 2$ which is a legitimate technique for this integrand. There are a few things to take note of:

1. When we make a change of variables we must also remember to change the limit of integration. For our substitution, $u=x\ln 2$, when $x=0\Rightarrow u=0$ & when $x=2\Rightarrow u=2\ln 2$. These now become your upper and lower limits, respectively. In other words our new definite integral to solve is

$\frac{1}{\ln 2}\int_{0}^{2\ln 2} e^{u}+6 du$,

and we don't need to go back to the x-variable.

2. On the second line of your work you switch back to $x.$ You don't need to do this if you follow the u-substitution all the way through. Also, even if we wanted to switch back to $x$ we must be sure to replace du properly too. It looks as though when you switched back to x, you replaced du with dx, when really $du=dx\ln 2.$

Does this help clear everything up?

Again, nice work and good luck!