Basic Calculus II Integral Questions - Please Help

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ardentmed

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Hey guys, I need some quick help for a. and b. (ignore the last two)
2014_07_14_369_be38da7c84555c6e2b50_2.jpg

I'm confused as to how these work. For instance, I'm assuming that the FTC must be used, where the formula used would be f(b)-f(a), correct? :confused:
If so, I got 4x^3 radical(2-x^4) -2 for the first one. The second one is 3x^2 (sin(x^3)) + sincosx * cosx.



Thanks in advance, guys. I really appreciate it.
 
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ardentmed said:
2014_07_14_369_be38da7c84555c6e2b50_1.jpg

The u-substitution for the first one is somewhat tricky. I ended up getting 1/40(u)^5/2 - 2 (u) ^3/2 +C, which I'm not too sure about. I took u from radical 3+2x^4.

For the second question, I split the integral into two parts, one from 3 to 1, and the other from 1 to -2, and used a negative value for the latter, since x < 1 is negative. This gave me 20/3 as a final answer.

Moreover, for the third question, I used a trigonometric identity u substitution and got pi^2 / 72.

For the last question (the word problem), I got -3.33m for displacement and took the absolute for velocity to get speed, which gave me 218/3 as the final answer.

Also, for these questions:
2014_07_14_369_be38da7c84555c6e2b50_2.jpg

I'm confused as to how these work. For instance, I'm assuming that the FTC must be used, where the formula used would be f(b)-f(a), correct? :confused:
If so, I got 4x^3 radical(2-x^4) -2 for the first one. The second one is 3x^2 (sin(x^3)) + sincosx * cosx.

Also for the last two, I got 3/5x^5/4 -2x + 3x^1/3 +c from multiplying the values in the bracket by themselves (expansion). Also, I got 2 for the final one, since sinx > 0 is positive, and sinx < 0 is negative, although I'm not too sure about that.


Thanks in advance, guys. I really appreciate it.
Click to expand...

Please refer to your first post:

http://www.freemathhelp.com/forum/threads/87512-Basic-Calculus-II-Help?p=359255#post359255

Please post 1 problem in one thread.
 
For \(\displaystyle \int x^7\sqrt{3+ 2x^4}dx\) I would, just to get rid of that complicated square root, let \(\displaystyle u= 3+ 2x^4\). Then \(\displaystyle du= 8x^3dx\) or \(\displaystyle \frac{1}{8}du= x^3dx\). We can write \(\displaystyle x^7= (x^4)(x^3)\) so that the integral is \(\displaystyle \int x^4\sqrt{3+ 2x^4}(x^3dx)= \int x^4\sqrt{u}du\). We still need to get rid of that "\(\displaystyle x^4\). If \(\displaystyle u= 3+ 2x^4\) then \(\displaystyle x^4= (u- 3)/2\) so the integral becomes \(\displaystyle \int (u- 3)\sqrt{u}/2 du\)\(\displaystyle = \int (u- 3)u^{1/2}/2 du\)\(\displaystyle = \frac{1}{2}\int u^{3/2}- 3u^{1/2} du\)

For \(\displaystyle \int_{-2}^3|x^2- 1|dx\) start by recognizing that \(\displaystyle x^2- 1= (x- 1)(x+ 1)\).

If x< -1, both x- 1 and x+ 1 are negative so \(\displaystyle x^2- 1\) is the product of two negative numbers. it is positive and \(\displaystyle |x^2- 1|= x^2- 1\).

If -1< x< 1, x- 1 is negative and x+ 1 is positive so \(\displaystyle x^2- 1\) is the product of a positive and a negative number. It is negative and \(\displaystyle |x^2- 1|= -(x^2- 1)= 1- x^2\).

Finally, if x> 1, both x- 1 and x+ 1 are positive so \(\displaystyle x^2- 1\) is the product of two positive numbers. It is positive and \(\displaystyle |x^2- 1|= x^2- 1\).

So \(\displaystyle \int_{-2}^3 |x^2- 1| dx= \int_{-2}^{-1} x^2- 1 dx+ \int_{-1}^1 1- x^2 dx+ \int_{1}^3 x^2- 1 dx\).
 
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You say "(a) and (b) are the only ones I don't understand" but you have two different set of problems labeled "a" and "b". Which do you mean?
 
Hi there. If you take pictures of exercises, instead of typing them, please crop your images appropriately before posting. Thank you! :cool:

You may type integrals using ASCII characters.

f(x) = INT[x^4 .. -2] sqrt(2-u) du

g(x) = INT[x^3 .. cos(x)] sin(u) du
 
The first of those, (a) \(\displaystyle \int_{-2}^3 |x^2- 1|dx\) I did above.

For (b) \(\displaystyle \int_0^{1/2} \frac{sin^{-1}(x)}{1- x^2} dx\), try the substitution \(\displaystyle u= sin^{-1}(x)\). What is du?
 
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