Question

cal 1 question

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

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.

Answer 5

then how do u do it with that limit?

Answer 6

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

Answer 7

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

Answer 8

But that's not the correct answer to it!

Answer 9

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.

Answer 10

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

Answer 11

answer should be 1027?

Answer 12

Nope :(

Answer 13

what is the answer? is it -1023?

Answer 14

Not even that..

Answer 15

answer is?

Answer 16

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

Answer 17

okay

Answer 18

did either work?

Answer 19

no