Suppose int[3,0][f(x+k)]dx = 4, for k constant. Then....

[paulxzt]

Can someone help me with this?

Suppose that the integral from 3 to 0 of f(x+k)dx = 4 where k is a constant. Then the integral from (3+k) to k of f(x)(dx) equals?

I integrated the first one and got

(1/2)x^2 + kx from 3 to 0

k = -1/6

Plugged it into the second. Integral from 17/6 to -1/6 of f(x)dx. what do I do now? do i just subtract the two?

 

[skeeter]

the integral from 3 to 0 of f(x+k)dx = 4 where k is a constant

\(\displaystyle \L \int_3^0 f(x+k)dx = 4\)

\(\displaystyle \L u = x+k\)
\(\displaystyle \L du = dx\)

substitute ...

\(\displaystyle \L \int_{3+k}^k f(u) du = 4\)

so ... \(\displaystyle \L \int_{3+k}^k f(x) dx = ?\)

 

[paulxzt]

sorry it was supposed to be from 0 to 3 and k to 3+k..

so is it just the integral from k to 3+k of (x+k) =4 ?

 

[pka]

It really worries me that your reply shows that you do not have a clue about what is question means. We know that because the minor change in notation has absolutely no effect on the answer.

 

[paulxzt]

It really worries me that your reply shows that you do not have a clue about what is question means. We know that because the minor change in notation has absolutely no effect on the answer.

I'm sorry for not grasping the concept so quickly? I was only able to sit in for one class lecture and it was the introduction to integrals. I was sick for a week but got the work from a friend to complete over the break

 

[pka]

That is fair enough. Here is the point of the question.

\(\displaystyle \L\int\limits_a^b {f(t)dt} = \int\limits_a^b {f(u)du} = \int\limits_{a - k}^{b - k} {f(x + k)dx} .\)