Question

Which of the following is closest to
[integrand] (upper limit is 3 & lower limit is 2) ((x^4)/2 )- (4e^x) dx?
A: -29.7
B: -18.3
C: 78.3

Answer 1

I'm guessing we can use a calculator to evaluate the limit then.

Answer 2

@Brutal It's an integral :-)

Answer 3

http://www.wolframalpha.com/input/?i=Integral+%28%28%28x%5E4%29%2F2%29-+%284e%5Ex%29%2Cx%2C2%2C3%29
This is about -29.6

Answer 4

Oh yea, I meant integral, those limits threw me off. @lanad8 the official notation would be:
Int(f(x),x,a,b)
where a is the lower limit, and b is the upper limit, so in this case:
Int( ((x^4)/2)- (4e^x), x, 2, 3)

Answer 5

I took the integral and found 1x^5/10 -4e^x. What did you get at least for the integral without evaluating it?

Answer 6

Actually this is real easy to integrate:
\[\int((x^4)/2- (4e^x))dx = \]
\[(1/2)*\int x^4dx - 4\int e^xdx=\]
\[(1/2)*(x^5)/5-4e^x+C=\]
\[x^5/10-4e^x+C\]

Answer 7

that's without evaluating the integral at the limits.

Answer 8

I evaluated it from 2 to 3 and got -29.7

Answer 9

ok, thanks brinethery & Brutal!

Answer 10

yvm!

May I recommend theintegralcalc's channel? She's good at throwing tough examples out :-)

Answer 11

i'll check out her channel. thank you :)

Answer 12

http://www.youtube.com/watch?v=DUowajlKIKA&feature=plcp&context=C45fb838VDvjVQa1PpcFOXN12qzTsROoA1BHwjds91n2gB--v-Gss%3D

Answer 13

http://www.youtube.com/user/TheIntegralCALC/videos?query=integral

Answer 14

PatrickJMT rules on youtube.

Answer 15

Hmm... she had the integral of 1/x^5. Wouldn't the antiderivative be ln|x^5|?

Answer 16

First of all no, she had the integral of \[x^5\]
Second of all, no the antiderivative of \[1/x^5\] is not \[ln|x^5|\]
\[\int(1/x^5)dx=\int(x^{-5})dx=x^{-5+1}/(-5+1)+C \]

Answer 17

Or:
\[-x^{-4}/4+C\]

Answer 18

But then what's the derivative of ln(x)?

Answer 19

\[(d/dx)(ln(x))=(1/x)\]

Answer 20

See?!

Answer 21

lol :-)

Answer 22

\[(d/dx)(ln(x^4))=(1/(x^4))*4x^3=4x^3/x^4=4/x\]

Answer 23

The only time you use Ln is when you have \[1/x\]because if you try to integrate that the proper way, you'll get a 0 in the denominator.

Answer 24

\[\int(1/x)dx=ln|x|+C\]
you can't do this:
\[\int(1/x)dx=\int(x^{-1})dx=x^{-1+1}/(-1+1)+C\]

Answer 25

Oh I see what you're saying.

Answer 26

See the bottom only gives you trouble with the power of -1. Otherwise you're good.

Answer 27

It's been like... 6 years since I've taken calc so I've forgotten a lot hehe

Answer 28

Thanks for explaining that :-)

Answer 29

np