Rename the inside to undo the chain rule

Mathematics · Calculus & Analysis · ages 19-20

Name ______________________ Date ____________

You want to find the integral of 2x cos(x^2) dx. What is the best choice for u?
- u = cos(x^2)
- u = x^2
- u = 2x
- u = x
You want the integral of 2x cos(x^2) dx. What is the best choice for u?
- u = cos(x^2)
- u = x^2
- u = 2x
What is the integral of x e^(x^2) dx?
- e^(x^2) + C
- x e^(x^2) + C
- (1/2) e^(x^2) + C
If u = x^2 + 4, what is du?
- du = 2x dx
- du = dx
- du = x^2 dx
For the definite integral from 1 to 2 of x(2 + x^2)^5 dx, what are the correct new limits after choosing u = 2 + x^2?
- u = 1 and u = 2
- u = 3 and u = 6
- u = 0 and u = 4
- u = 2 and u = 2
Evaluate the definite integral from 0 to 1 of 2x(1 + x^2)^3 dx. Give the exact value as a decimal.
Answer: ______________
Why does the substitution u = x^2 fail for the integral of cos(x^2) dx?
- Because cos is not allowed in a substitution
- Because du = 2x dx needs a factor of x that is not in the integrand
- Because the integral must be definite first
- Because u must always be x
Which substitution makes the integral of x / (1 + x^2) dx work?
- u = x, because it is the simplest variable
- u = 1 + x^2, because du = 2x dx absorbs the x on top
- u = x^2, because the denominator is squared
- u = 1 / (1 + x^2), because it is the denominator
Why can the integral of e^(x^2) dx not be finished with u = x^2?
- Exponentials reject every substitution
- The limits were never converted
- du needs a factor of x that is missing
Evaluate the definite integral from 1 to e of (ln x) / x dx, where e is Euler's number (about 2.718). Give the exact value as a decimal.
Answer: ______________

Answer key

For grown-ups. Fold this page away before handing over the rest.

Rename the inside to undo the chain rule W1-mt_jH-j_XsdC7-s1

u = x^2 · Set u equal to the inner expression x^2, because its derivative 2x is already sitting in the integrand.
u = x^2 · The inner expression is x^2, and its derivative 2x is already in the integrand.
(1/2) e^(x^2) + C · With u = x^2, a factor of 1/2 adjusts for the missing 2, giving (1/2) e^(x^2) + C.
du = 2x dx · Differentiate x^2 + 4 with respect to x to get 2x, so du = 2x dx.
u = 3 and u = 6 · Plug the old limits into u = 2 + x^2: x = 1 gives u = 3, and x = 2 gives u = 6.
3.75 · With u = 1 + x^2 the limits become u = 1 and u = 2, and the integral becomes the integral from 1 to 2 of u^3 du, which is 15/4 = 3.75.
Because du = 2x dx needs a factor of x that is not in the integrand · Substitution only works when du absorbs the leftover factor. Here du = 2x dx, but there is no x in the integrand to absorb it.
u = 1 + x^2, because du = 2x dx absorbs the x on top · Set u = 1 + x^2. Then du = 2x dx, and the numerator x dx supplies exactly half of du, leaving a simple integral of 1/u.
du needs a factor of x that is missing · Substitution only works when du absorbs the leftover factor, and here nothing supplies the x.
0.5 · Choose u = ln(x). The limits become u = 0 and u = 1, and the integral of u du from 0 to 1 equals 1/2.