Evaluate the following integral:
(4-2x)dx [2,5]
I can get it part way then I freeze.

Question

Evaluate the following integral:
(4-2x)dx  [2,5]
I can get it part way then I freeze.

nubeer · Answer

$$\int\limits_{2}^{5} 4 -\int\limits_{2}^{5} 2x$$
can you do now?

anonymous · Answer

$$\int\limits_{2}^{5} 4(dx) = 4x  -   2\int\limits_{2}^{5} x(dx) = x^{2}?????$$
that is where I get up to and now I am stuck

nubeer · Answer

hmm let me remind u the rule
∫x^n = [x^(n+1)]/ (n+1)

nubeer · Answer

there is 1 added in the power.. and its divided by the new power we got.

anonymous · Answer

so x^2/2 right?

nubeer · Answer

yes. :)

anonymous · Answer

So do I then do (5-2)^2/2?

nubeer · Answer

hmm? how you got 5 here?
its like 4x - x^2

nubeer · Answer

and aplly the limit on this.

anonymous · Answer

so 4(5-2) - (5-2)^2?

anonymous · Answer

nope, wrong answer.

anonymous · Answer

the limit is [2,5] right?

nubeer · Answer

hmm no.. that is not how we apply limit..
[4(5)-(5)^2] - [ 4(2)-(2)^2]

anonymous · Answer

ahhhhhh ..... grrrrrr.
so this equals -9.

nubeer · Answer

yes . i think so.. is it wrong?

anonymous · Answer

nope that is what the back of the book says. What ever happened to that 2 that was before the integral?

nubeer · Answer

hmm no look
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