How Do I Solve These Tricky Integrals Using Integration by Parts?

[VADER25]

(2x+7)squareroot(7-x)dx from x=6 to x=7

i've been trying to solve this integral with the integration by parts formula but it won't work.
would appreciate some help.

would like to see the whole solution and not just the answer. (the answer should bee in fractions)

 

[G01]

would like to see the whole solution and not just the answer. (the answer should bee in fractions)

I would like to see some work...You have to show some work in order to get help here. Let us see what you have tried so far.

 

[VADER25]

(2x+7)squareroot(7-x)dx from x=6 to x=7

u=squareroot(7-x)

u^2=(7-x)
x=7-u^2

dx=-2 du

(2x+7)squareroot(7-x)dx

(2(7-u^2)+7)u dx


(14-2u^2+7)u dx

(21-2u^2)u * -2u


(21-2u^3)*-2u


dont know how to continue from here.

 

[asd1249jf]

(2x+7)squareroot(7-x)dx from x=6 to x=7

u=squareroot(7-x)

u^2=(7-x)
x=7-u^2

dx=-2 du

(2x+7)squareroot(7-x)dx

(2(7-u^2)+7)u dx


(14-2u^2+7)u dx

(21-2u^2)u * -2u


(21-2u^3)*-2u


dont know how to continue from here.

You aren't performing integration by parts to solve this problem, from what I can see.

http://en.wikipedia.org/wiki/Integration_by_parts

Read that article.

 

[G01]

(2x+7)squareroot(7-x)dx from x=6 to x=7

u=squareroot(7-x)

u^2=(7-x)
x=7-u^2

dx=-2 du

(2x+7)squareroot(7-x)dx

(2(7-u^2)+7)u dx(14-2u^2+7)u dx

(21-2u^2)u * -2u(21-2u^3)*-2udont know how to continue from here.

As l46kok said, what you are doing is not integration by parts. You are trying to integration by substitution from what I can tell.

Here is an article on integration by parts from Wolfram MathWorld:

http://mathworld.wolfram.com/IntegrationbyParts.html

 

[Avodyne]

(21-2u^3)*(-2u)
dont know how to continue from here.

I've put in some parentheses to make the expression unambiguous.

Now multiply out and integrate each term.

 

[VADER25]

i think i need to see a complete solution to this problem, I've been working on this for three days now and stiill don't understand it completely.

$$\int_{1}^{2}(9x+6)\sqrt{7-3x} dx\$$

 

[HallsofIvy]

(2x+7)squareroot(7-x)dx from x=6 to x=7

u=squareroot(7-x)

u^2=(7-x)
x=7-u^2

dx=-2 du

(2x+7)squareroot(7-x)dx

(2(7-u^2)+7)u dx


(14-2u^2+7)u dx

(21-2u^2)u * -2u

Well, (21- 2u^2)(-2udu)

(21-2u^3)*-2u


dont know how to continue from here.

Well, that's NOT integration by parts, it's a straight forward substitution. But it looks to me like you are doing fine. I don't see why you only multiplied one of the "u"s and not both. : (21- 2u^2)(-2u du)= (4u^3- 42u)du. You might want to change the limits of integeration also. When x= 6, what is u? When x= 7 what is u?

 

[VADER25]

i know now how to get it right, thanks for your help.

 

[VADER25]

i have a new problem, I've tried to solve it but i think i need some advice.

$$\int_{0}^{\pi /2}sin7xcos6x\hspace{6} dx\\ \\ f=cos6x\hspace{6} F=\frac{sin6x}{6}\\ g'=7cos7x\hspace{6} g=sin7x\\$$

and this is what i have done so far

$$\int_{0}^{\pi /2}sin7xcos6x\hspace{6} dx\\ \\ (\frac{1}{6}sin6x)(sin7x)-\int_{0}^{\pi /2}(\frac{1}{6}sin6x)(7cos7x)=\\ \\ (\frac{1}{6}sin6x)(sin7x)-\frac{7}{6}\int_{0}^{\pi /2}(sin6x)(cos7x)\hspace{6}$$