## ego_caelestis 3 years ago $\int\limits_{0}^{4}\sqrt{3x+4}=\int\limits_{4}^{12}u^{1/2}(1/3)du$ when substituting u=sqrt{3x+4} which is all good but what i don't get is (1/3)*[(2/3)u^(3/2)]|from 4 to 16 where does the 2/3 and u^(3/2) come from?

1. elica85

1/3 is a constant that can be pulled out of the integral. when using u sub, you need to plug in the limits so in this case, set u=3x+4. plug in the limits u=3(0)+4 and u=3(4)+4. you have the new integral with new limits and u is raised to 1/2 which is the same as saying sqrt(u)

2. elica85

1/3 come from differentiating u so you have du=3dx. solve for dx=1/3du

3. ego_caelestis

right i got that but i don't get where the 2/3*u^(3/2) comes from

4. elica85

i don't see 2/3. what you posted is 1/3 and u^1/2

5. ego_caelestis

in the rest of the text (1/3)*[(2/3)u^(3/2)}

6. elica85

o you mean the final answer? say you have to integrate x^2 dx. your answer is (x^2+1)/(2+1) right?

7. ego_caelestis

yeah

8. elica85

so without looking at 1/3 (since it gets pulled out) it will be [u^(1/2+1)]/(1/2+1) which is 2/3u^3/2

9. ego_caelestis

thank you i don't know why i wasn't seeing that

10. elica85

np, 2/3 on the bottom was flipped so gets hard to spot

11. ego_caelestis

i was starring at that for half an hour before i thought to come here and now you've made it worth it thank you.