• May 30th 2010, 11:49 PM
nuvydeep
Hello to my lovely fellow math help forum posters,

I am in need of help with the following problems involving evaluating integrals. I took a picture of the problems that I need help with. It's 3 integral problems and 1 derivative problem involving max/min etc.

http://sphotos.ak.fbcdn.net/hphotos-..._8089916_n.jpg

I would appreciate any help that can be given!!! I need help on the steps to this method.. step by step. Can someone write the steps out and take a picture of it and post it up because its easier than writing *int* instead of the integral sign? (Headbang)

Thanks (Nod)
• May 31st 2010, 12:01 AM
autumn
In (a) let $u=\cos x$, so $du=-\sin x dx$ you do the rest.

For (b) let $u=1+\ln x$, so $du=x^{-1}dx$

For (c) just do parts twice.
• May 31st 2010, 12:01 AM
Prove It
Quote:

Originally Posted by nuvydeep
Hello to my lovely fellow math help forum posters,

I am in need of help with the following problems involving evaluating integrals. I took a picture of the problems that I need help with. It's 3 integral problems and 1 derivative problem involving max/min etc.

http://sphotos.ak.fbcdn.net/hphotos-..._8089916_n.jpg

I would appreciate any help that can be given!!! I need help on the steps to this method.. step by step. Can someone write the steps out and take a picture of it and post it up because its easier than writing *int* instead of the integral sign? (Headbang)

Thanks (Nod)

We're not here to do your homework for you...

Just a few hints:

6. a) Make the substitution $u = \cos{x}$.

6. b) Make the substitution $u = \ln{x} + 1$.

6. c) You will need to use integration by parts twice.

7. a) The function is increasing when the derivative is positive, and decreasing when the derivative is negative.

7. b) The function is concave up when the derivative is increasing, i.e. when the second derivative is positive, and concave down when the derivative is decreasing, i.e. where the second derivative is negative. There are inflection points where it changes from being concave up to concave down, i.e. where the second derivative is 0 or does not exist.

In future, please show what work you have done and exactly where you need help.
• May 31st 2010, 12:09 AM
tonio
Quote:

Originally Posted by nuvydeep
Hello to my lovely fellow math help forum posters,

I am in need of help with the following problems involving evaluating integrals. I took a picture of the problems that I need help with. It's 3 integral problems and 1 derivative problem involving max/min etc.

http://sphotos.ak.fbcdn.net/hphotos-..._8089916_n.jpg

I would appreciate any help that can be given!!! I need help on the steps to this method.. step by step. Can someone write the steps out and take a picture of it and post it up because its easier than writing *int* instead of the integral sign? (Headbang)

Thanks (Nod)

Hint:

(1) For 1st integral: $\int f'(x)f^2(x)\,dx=\frac{f^3(x)}{3}+C\,,\,\,f(x)$ any derivable function

(2) For 2nd integral: if $\int f(x)\,dx=F(x)\,,\,\,then\,\,\,\int g'(x)f(g(x))\,dx=F(g(x))$ , with $g(x)$ a derivable function ( note that here $F(x)$ is the primitive function of $f(x)$ )

(3) For 3rd integral: do integration by parts with $u=x^2\Longrightarrow u'=2x\,,\,\,v'=e^{-x}\Longrightarrow v=-e^{-x}$

(4) for 4th problem: a twice derivable function $f(x)$ has:

(i) a maximum point at $(x_0,f(x_0))$ if $f'(x_0)=0\,,\,\,f''(x_0)<0$

(ii) a minimum point at $(x_0,f(x_0))$ if $f'(x_0)=0\,,\,\,f''(x_0)>0$

The function is increasing when $f'(x)>0$ and decreasing when $f'(x)<0$ , and concave upwards when $f''(x)>0$ and concave downwards when $f''(x)<0$ .

The function has an inflection point when the second derivative has different signs on the left and on the right of that point.

Tonio