Math Help - 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. $\int_2^5 4h(x)dx = \int_2^3 4h(x)dx + \int_3^5 4h(x) dx = 24 + \int_3^5 4h(x)dx$.

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

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 $h(x) = k(x)$, 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