Substitution
- Page ID
- 218534
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Advanced Substitution
Substituting x
Example Find \( \int x \sqrt{x-1} dx \)
Solution Notice first that the square root is what makes this problem difficult. Hence we let u = x - 1 du = dx This seems not to get rid of the x term, but notice that if we add 1 to both sides, we get x = u + 1 Now we can substitute to get \( \int (u + 1) \sqrt{u} du = \int u^{\frac{3}{2}} + u^{\frac{1}{2}} du \) = 2/5 u5/2 + 2/3 u3/2 + C = 2/5 (x - 1)5/2 + 2/3 (x - 1)3/2 + C
Exercises: Find the following integrals A. \( \int \frac{x}{1 - x} dx \) B. \( \int \frac{x^2}{1 - x} dx \) C. \( \int \frac{x^3}{1 - x} dx \)
Summary of Substitution A list of "good u's" follows: A. Inside the parentheses B. In the square root C. The denominator E. In the exponential F. Be creative
Definite Integrals We can use substitution to evaluate definite integrals. When using substitution, we substitute the values for u and do not resubstitute at the end.
Example Evaluate \( \int_0^1 \frac{x^2}{1 + x^3} dx \)
Solution This is a substitution with u = 1 + x3 du = 3x2 dx 1/3 du = x2 dx This is just old substitution business. The new part is how to deal with the limits 0 and 1. We just plug into x and find u. When x = 0 u = 1 + 03 = 1 When x = 1 u = 1 + 13 = 2 Now we can substitute all the numbers and expressions to get \( \frac{1}{3}\int_1^2 \frac{1}{u} du = [\frac{1}{3}ln|u|]_1^2\) Now we just plug in the numbers 2 and 1. Notice that there is no " + C" since it is a definite integral. We also do not need to resubstitute since the numbers already represent "u" values. Plugging in gives 1/3 ( ln 2 - ln 1) = 1/3 ln 2
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