## anonymous 5 years ago cal 1 question

1. anonymous

2. anonymous

okay so you can separate the integral like $\int\limits_{-5}^{1}f(x) dx= \int\limits_{-5}^{0}82xdx+ \int\limits_{0}^{1}6x^2dx$

3. anonymous

technically you would have to evaluate the lower limit of the integral of the second term ( the 0 in the 6x^2 integral ) at some letter (a,b,c ect what ever letter you like). then you take the limit of the letter as it approaches the zero that is there as you see it above.

4. anonymous

So in a more punctual approach the integral should be $\int\limits_{1}^{-5}f(x) dx = \int\limits_{-5}^{0}82x dx + \lim_{a \rightarrow 0} \int\limits_{a}^{1}6x^2dx$

5. anonymous

It's kind of technicality you will definitely see in later calculus called improper integrals. The REASON for this is because of the inequality $x > 0$ which means the x on that interval never actually reaches zero, so you say as x --> 0 (as x approaches zero, 0.00000001 and smaller). Now just take the integral of either one, the second one is the most correct if you understand the limit deal. but both will give you the same answer.

6. anonymous

then how do u do it with that limit?

7. anonymous

you would evaluate the integral first then take the limit. so $\lim_{a \rightarrow 0} \int\limits_{a}^{1} 6x^2dx = \lim_{a \rightarrow 0} 6x^3/3- a \rightarrow1$

8. anonymous

then $6(1)^3/3 - \lim_{a \rightarrow 0}6(a)^3/3 = 2(1)^3 -\lim_{a \rightarrow 0}2(a)^3 = 2 - 0 = 2$

9. anonymous

But that's not the correct answer to it!

10. anonymous

thats only half of it though. you have to take the integral of the first half also, i was only showing how to do the limit part.

11. anonymous

do you see? so evaluate $\int\limits_{-5}^{0}82xdx$ and add two

12. anonymous

13. anonymous

Nope :(

14. anonymous

what is the answer? is it -1023?

15. anonymous

Not even that..

16. anonymous

17. anonymous

I can just type in the possible answers to see if thats right or wrong.. It doesn't tell the exact ans though

18. anonymous

okay

19. anonymous

did either work?

20. anonymous

no