# Thread: Stuck - Integration

1. ## Stuck - Integration

I just try but i don't find any relation for the attached question.

sorry if this question not under this section.

2. ## Re: Stuck - Integration

Use what you know. $\displaystyle \int_2^5 4h(x)dx = \int_2^3 4h(x)dx + \int_3^5 4h(x) dx = 24 + \int_3^5 4h(x)dx$.

So, for part (a), the answer is $\displaystyle 24 + 4\int_3^5 h(x)dx$.

Do something similar for part (b). You won't be able to get rid of the integral entirely, but you can limit the bounds of integration.

3. ## Re: Stuck - Integration

tq for the respond mr slip

that what i try to do...but before i start the h(x) & k(x) make me stuck...when i check the formulae....the function should be same h(x) or k(x) then we can start the integration....correct me if wrong

4. ## Re: Stuck - Integration

If $\displaystyle h(x) = k(x)$, then yes, you could determine values for each part that did not include integrals.

5. ## Re: Stuck - Integration

Dear SlipEternal..

I agree with your Q(a) ans...

but i dont agree with the answer given on the book as attached. i think the answer is wrong both for (a) & (b)

6. ## Re: Stuck - Integration

Dear All..

I just update my calculation base on h(x) = k(x) ..i get the ans for (a) 64 but not for (b)..hope u have time to confirm my calculation on (b)..please find attached

7. ## Re: Stuck - Integration

Originally Posted by nikconsult
Dear All..

I just update my calculation base on h(x) = k(x) ..i get the ans for (a) 64 but not for (b)..hope u have time to confirm my calculation on (b)..please find attached
If the functions are equal, then your answers are correct.

8. ## Re: Stuck - Integration

Originally Posted by SlipEternal
If the functions are equal, then your answers are correct.
tq for your respond SlipEternal

do you mean my ans for (b) section is correct as per my attachement??