# Integration

• Apr 18th 2009, 02:05 PM
student_forever
Integration
Could you explain the way to perform the following integrations, please?
(see the attachment)

Thank you!!!!!!!
• Apr 18th 2009, 02:57 PM
Calculus26
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 PM
mr fantastic
Quote:

Originally Posted by student_forever
Could you explain the way to perform the following integrations, please?
(see the attachment)

Thank you!!!!!!!

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:

$\int_{-1}^1 1 + e^x \, dx$
Each of these terms in the integrand is a standard form. Refer to the tables here: Lists of integrals - Wikipedia, the free encyclopedia

Quote:

$\int_0^4 x^3 \sqrt{x^4 + 1} \, dx$
Make the substitution $u = x^4 + 1$.
• Apr 19th 2009, 09:55 AM
student_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 AM
skeeter
Quote:

Originally Posted by student_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 :(

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

substitute ...

$\frac{1}{4} \int_1^{257} \sqrt{u} \, du$

now integrate.
• Apr 19th 2009, 11:34 AM
student_forever
Sorry, guys, I just dont get it(Headbang)
PLEASE, can you show me the solution step by step??
• Apr 19th 2009, 11:52 AM
derfleurer
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.

• Apr 19th 2009, 11:56 AM
skeeter
Quote:

Originally Posted by skeeter
$\frac{1}{4} \int_0^4 4x^3\sqrt{x^4+1} \, dx$

substitute ...

$\frac{1}{4} \int_1^{257} \sqrt{u} \, du$

now integrate.

$\frac{1}{4}\left[\frac{2}{3}u^{\frac{3}{2}}\right]_1^{257}$

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

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

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

$\frac{1}{6}\left[(257)^{\frac{3}{2}} - 1\right]$
• Apr 19th 2009, 01:10 PM
student_forever
Skeeter, can you please write dow the solution to problem J?(Crying)
• Apr 19th 2009, 02:10 PM
student_forever
never mind, I got it!!

Thanx to all of you!!!