Could you explain the way to perform the following integrations, please?

(see the attachment)

Thank you!!!!!!!

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- Apr 18th 2009, 02:05 PMstudent_foreverIntegration
Could you explain the way to perform the following integrations, please?

(see the attachment)

Thank you!!!!!!! - Apr 18th 2009, 02:57 PMCalculus26
In J you have a straight forward anti derivative x+e^x

In K use u - substitution

let u = x^4 +1 the anti-derviative is then simple - Apr 18th 2009, 03:18 PMmr fantastic
Please take the small time to type your questions. Perhaps even learn some basic latex, see here: http://www.mathhelpforum.com/math-help/latex-help/.

Quote:

$\displaystyle \int_{-1}^1 1 + e^x \, dx$

Quote:

$\displaystyle \int_0^4 x^3 \sqrt{x^4 + 1} \, dx$

- Apr 19th 2009, 09:55 AMstudent_forever
for K:

I have u=x^4+1 , du=4x^3, dx=1/4x^3

when x=4, u=257; when x=0, u=1

so now I have \intx^3\sqrt{u}*1/4x^3 correct?

what do I do next?

P.S. I am trying to write it in latex style, it doesn't work :( - Apr 19th 2009, 10:14 AMskeeter
- Apr 19th 2009, 11:34 AMstudent_forever
Sorry, guys, I just dont get it(Headbang)

the answer should be 1/6[(257)^3/2-1)~686.50

PLEASE, can you show me the solution step by step?? - Apr 19th 2009, 11:52 AMderfleurer
What's the derivative of u^n? It's n*u^(n-1)*u', right? (just chain rule, bring the exponent out front, take one from that exponent, multiply by the derivative of the inside) Like, the derivative of (x^3 + 2)^2 is 2(x^3 + 2)(3x^2).

When integrating, we just tak the opposite approach. That is to say, we raise n by 1, divide by n + 1, and then divide by the derivative of the inside.

Look again at 2(x^3 + 2)(3x^2). Let's see if we can get back our original (x^3 + 2)^2. Let u = x^3 + 2. Since we're working backward, we know the form n*u^(n-1)*u' has an integral of u^n. Our (x^3 + 2) is being raised to the power of 1, right? So when does n-1=1? When n is 2. This means that so far, we're in the correct form n*u^(n-1). However, we haven't accounted for u'. If we are infact multiplying n*u^(n-1) by u', than we know our answer is just u^n. Lucky us, (3x^2) is u'. So it's in the correct form to convert to u^n, where n is 2 and u is x^3 + 2.

Same concept with x^3(x^4 + 1)^(1/2). Let u=x^4 + 1. Our n is then 1/2 + 1 (or 3/2). Next, however, we don't have a 3/2 coefficient, do we? We have a coefficient of 1. So, multiply by 2/3 to account for that. There's just one other thing to check before we convert to u^n, our u'. Unfortunately again, we're multiplying by x^3, not 4x^3. But, like before, we just multiply by 1/4 to balance it out.

Thus you get your answer of (2/3)(1/4)u^(3/2) or (1/6)u^(3/2) - Apr 19th 2009, 11:56 AMskeeter
$\displaystyle \frac{1}{4}\left[\frac{2}{3}u^{\frac{3}{2}}\right]_1^{257}$

$\displaystyle \frac{2}{12}\left[u^{\frac{3}{2}}\right]_1^{257}$

$\displaystyle \frac{1}{6}\left[u^{\frac{3}{2}}\right]_1^{257}$

$\displaystyle \frac{1}{6}\left[(257)^{\frac{3}{2}} - 1^{\frac{3}{2}}\right]$

$\displaystyle \frac{1}{6}\left[(257)^{\frac{3}{2}} - 1\right]$ - Apr 19th 2009, 01:10 PMstudent_forever
Skeeter, can you please write dow the solution to problem J?(Crying)

- Apr 19th 2009, 02:10 PMstudent_forever
never mind, I got it!!

Thanx to all of you!!!