## monroe17 2 years ago someone help? (:

1. monroe17

$\int\limits_{-2}^{0} (\frac{ 1 }{ 2 }t^4+\frac{ 1 }{ 4 }t^3-t)dt$

2. nubeer

Hint. $\int\limits_{?}^{?} t ^{4} =\frac{ t ^{4+1} }{ 4+1}$

3. monroe17

I honestly don't even know how to solve or begin to solve for it.. can you help me step by step? I have 7 more problems like this and I'd like to do them all myself.. But I need help solving and the steps to do so.

are you familiar with the general formula of integration?

5. monroe17

I'm familiar with... $\int\limits_{a}^{b} f(x)dx=F(x)|_{a}^{b} = F(b)-F(a)$

6. nubeer

ya this is done when we apply the limits but do you know how we do integration?

7. monroe17

I don't.

the general formula for integration of $$\int x^n dx = \frac{x^n+1}{n+1} +C$$

9. nubeer

|dw:1355140748339:dw|

10. winterfez

|dw:1355140890023:dw|

@winterfez i believe that's $$\left[ \frac{1}{2} \frac{t^5}{5}+\frac{1}{4} \frac{t^4}{4}- \frac{t^2}{2} \right]^0_{-2}$$

12. winterfez

$(\frac{ 1 }{ 2 }\frac{ 0 }{ 5 }+\frac{ 1 }{ 4 }\frac{ 0^{4} }{ 4 }-\frac{ 0 }{ 2 })-(\frac{ 1 }{ 2 }\frac{ (-2)^5 }{ 5 }+\frac{ 1 }{ 4 }\frac{ (-2)^4 }{ 4 }-\frac{ (-2)^{2} }{ 2 })$

13. winterfez

now use your calculator to punch in those suckers

14. monroe17

can you explain the [...]_{-2}^{0}

∫abf(x)dx=F(x)|ba=F(b)−F(a) <--this is why.

from your original post up there.

17. monroe17

ohhh so i do apply that.. I just missed a step.

yup :) i think you've got it.

19. monroe17

thank you guys! (: you're amazing!

you're welcome :)

21. monroe17

I got 21/5.. does that seem right @Shadowys

-21/5?

23. monroe17

err... let me recheck my work.

sorry, my mistake. i typed wrong. lol no neg.

25. monroe17

oh okay (: is that the answer though? because what happens with the .... +C?

well. that's for indefinite integrals because you don't know where is the limit. the C is gone when it is definite.

27. monroe17

oh okay (: so 21/5 it is?

yup! :)

29. monroe17

thanks again!