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 h00zah Feb2-11 11:15 PM

definite integration by parts with sub

hello, i am stuck on how to do this

I know how to do it for an indefinite integral, but it gets confusing for a definite integral. from my knowledge, when doing a definite integral, you have to change the upper and lower limit. but when it comes to integration by parts for a definite integral using substitution i get completely lost. my prof made an example using a indefinite integral, but not a definite integral.

is there a methodology to solving these?

 Mark44 Feb3-11 01:18 AM

Re: definite integration by parts with sub

Quote:
 Quote by h00zah (Post 3117670) hello, i am stuck on how to do this I know how to do it for an indefinite integral, but it gets confusing for a definite integral. from my knowledge, when doing a definite integral, you have to change the upper and lower limit.
If you do the integration by making a substitution you can either 1) change the limits of integration according to the substitution and evaluate the antiderivative at the changed limits of integration; OR 2) leave the limits unchanged, get the antiderivative, undo the substitution, and then evaluate. In other words, you don't have to change the limits of integration.
Quote:
 Quote by h00zah (Post 3117670) but when it comes to integration by parts for a definite integral using substitution i get completely lost. my prof made an example using a indefinite integral, but not a definite integral. is there a methodology to solving these?
To the side as scratchwork, do the indefinite integral and find the antiderivative. When you have found the antiderivative (and have undone any substitutions you might have done), evaluate your antiderivative at the two original limits of integration.

Here's an example that is done both ways, using an ordinary substitution:
$$\int_1^2 2x(x^2 + 1)^3 dx$$

1. Limits of integration unchanged
u = x2 + 1, du = 2xdx
$$\int_1^2 2x(x^2 + 1)^3 dx = \int_{x = 1}^2 u^3 du = \left.\frac{u^4}{4}\right|_{x = 1}^2$$
$$= \left.\frac{(x^2 + 1)^4}{4}\right|_{x = 1}^2 = \frac{625}{4} - \frac{16}{4} = \frac{609}{4}$$

2. Limits of integration changed per substitution
u = x2 + 1, du = 2xdx
$$\int_1^2 2x(x^2 + 1)^3 dx = \int_{u = 2}^5 u^3 du = \left.\frac{u^4}{4}\right|_{u = 2}^5$$
$$= \frac{625}{4} - \frac{16}{4} = \frac{609}{4}$$

In #2, when x = 1, u = 12 + 1 = 2,
and when x = 2, u = 22 + 1 = 5

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