Covers the two methods for handling the limits of integration when evaluating definite integrals with u-substitution: changing the limits to be in terms of u, or back-substituting and using the original limits.
What are the two valid methods for evaluating a definite integral \(\int_a^b f(g(x))g'(x) \, dx\) using u-substitution?
1. Change the limits of integration to be in terms of \(u\) and evaluate the new integral. \(\int_{g(a)}^{g(b)} f(u) \, du\).
2. Solve the indefinite integral first, back-substitute to get the answer in terms of \(x\), and then evaluate using the original limits \(a\) and \(b\).
Consider \(\int_0^2 x(x^2+1)^3 \, dx\). If we let \(u = x^2+1\), what do the limits of integration \(x=0\) and \(x=2\) become in terms of \(u\)?
For the lower limit: when \(x=0\), \(u = 0^2+1 = 1\). For the upper limit: when \(x=2\), \(u = 2^2+1 = 5\). The new limits are from \(1\) to \(5\).
Using the new limits \(u=1\) to \(u=5\), rewrite the integral \(\int_0^2 x(x^2+1)^3 \, dx\) completely in terms of \(u\).
With \(u=x^2+1\), we have \(du=2x\,dx\), so \(x\,dx = \frac{1}{2}du\). The integral becomes \(\int_1^5 u^3 \cdot \frac{1}{2}du = \frac{1}{2} \int_1^5 u^3 \, du\).
What is the main advantage of changing the limits of integration when using u-substitution for definite integrals?
It eliminates the need for the final back-substitution step. Once the integral is evaluated with the new \(u\)-limits, the calculation is complete, which can be faster and less error-prone.
Evaluate \(\int_0^{\pi/4} \sec^2(x) e^{\tan(x)} \, dx\) by changing the limits of integration.
Let \(u = \tan(x)\), so \(du = \sec^2(x) \, dx\). The lower limit is \(u = \tan(0) = 0\). The upper limit is \(u = \tan(\pi/4) = 1\). The integral becomes \(\int_0^1 e^u \, du = [e^u]_0^1 = e^1 - e^0 = e - 1\).
Now, evaluate \(\int_0^2 x(x^2+1)^3 \, dx\) using the second method (back-substituting before evaluating).
First, find the indefinite integral: \(\int x(x^2+1)^3 \, dx = \frac{1}{2} \int u^3 \, du = \frac{1}{8}u^4 + C = \frac{1}{8}(x^2+1)^4 + C\). Now evaluate using the original limits: \(\left[ \frac{1}{8}(x^2+1)^4 \right]_0^2 = \frac{1}{8}( (2^2+1)^4 - (0^2+1)^4 ) = \frac{1}{8}(5^4 - 1^4) = 78\).
For the integral \(\int_1^e \frac{\ln(x)}{x} \, dx\), what are the new limits if you set \(u = \ln(x)\)?
Lower limit: When \(x=1\), \(u = \ln(1) = 0\). Upper limit: When \(x=e\), \(u = \ln(e) = 1\). The new limits are from \(0\) to \(1\).
Solve \(\int_1^e \frac{\ln(x)}{x} \, dx\).
With \(u=\ln(x)\), \(du = \frac{1}{x}dx\), and limits \(0\) to \(1\). The integral is \(\int_0^1 u \, du = \left[ \frac{u^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}\).
