cal 1 question

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

cal 1 question

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

So in a more punctual approach the integral should be \[ \int\limits_{1}^{-5}f(x) dx = \int\limits_{-5}^{0}82x dx + \lim_{a \rightarrow 0} \int\limits_{a}^{1}6x^2dx\]
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.
then how do u do it with that limit?
you would evaluate the integral first then take the limit. so \[\lim_{a \rightarrow 0} \int\limits_{a}^{1} 6x^2dx = \lim_{a \rightarrow 0} 6x^3/3- a \rightarrow1\]
then \[6(1)^3/3 - \lim_{a \rightarrow 0}6(a)^3/3 = 2(1)^3 -\lim_{a \rightarrow 0}2(a)^3 = 2 - 0 = 2\]
But that's not the correct answer to it!
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.
do you see? so evaluate \[\int\limits_{-5}^{0}82xdx\] and add two
answer should be 1027?
Nope :(
what is the answer? is it -1023?
Not even that..
answer is?
I can just type in the possible answers to see if thats right or wrong.. It doesn't tell the exact ans though
okay
did either work?
no

Not the answer you are looking for?

Search for more explanations.

Ask your own question