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Substitution in definite integrals

Consider the following definite integral:

$\displaystyle \int_0^3 (1 + t^2)^4 tdt.$

We can do this by first doing the indefinite integral:

$\displaystyle \int (1 + t^2)^4 t dt = \int u^4 \frac{1}{2}du = \frac{1}{2}\int u^4 du = \frac{1}{2}\frac{u^5}{5} + C$

$\displaystyle = \frac{1}{10}u^5 + C = \frac{1}{10}(1 + t^2)^5 + C.$

Here we made the substitution $ u = 1 + t^2$, $ du = u' dt = 2t dt$.
Then by FTCII we have:

$\displaystyle \int_0^3 (1 + t^2)^4 tdt = [\frac{1}{10}(1 + t^2)^5]_0^3 = \frac{1}{10}(10^5 - 1) = 10000 - 0.1 = 9999.9$

Notice that when we make the substitution $ u = 1 + t^2$, when $ t = 0$, we have $ u = 1$ and when $ t = 3$, we have $ u = 10$.
Then we find:

$\displaystyle \int_1^{10} u^4 \frac{1}{2} du = [\frac{1}{10} u^5]_1^{10} = \frac{1}{10}(10^5 - 1) = 9999.9.$

So we get the right answer, when we make a substitution, if, when we introduce a new variable, we also change the limits of integration to the appropriate limits for the new variable.
This can be very useful, when it is necessary to make multiple substitutions, or if it is inconvenient to back substitute to the old variables, after the indefinite integral has been done.

We can see this in general as follows:
If $ F'(x) = f(x)$, then we have:

$\displaystyle \int_a^b f(x(t)) x'(t) dt = \int_a^b F'(x(t))x'(t) dt = \int_a^b \frac{d}{dt}(F(x(t))) = [F(x(t))]_a^b$

$\displaystyle = F(x(b)) - F(x(a)) =
\int_{x(a)}^{x(b)} F'(x)dx = \int_{x(a)}^{x(b)}f(x) dx.$

So we can convert a definite integral in the variable $ t$ into another equal definite integral, in the variable $ x$, provided that: Some examples:

$\displaystyle \int_0^3 t^2 (1 + t^3)^4 dt = \int_1^{28} \frac{1}{3} x^4 dx = \frac{1}{15}[x^5]_1^{28} = \frac{1}{15}(28^5 - 1) = \frac{5736789}{5}.$

Here we made the substitutions:

$\displaystyle \int_1^{9}\frac{dt}{\sqrt{2t + 7}} = \int_4^{25} u^{-\frac{1}{2}}\frac{1}{2} du = [u^{\frac{1}{2}}]_9^{25} = \sqrt{25} - \sqrt{9} = 2.$

Here we made the substitutions:

$\displaystyle \int_1^{e^2} \frac{\ln(t)}{t} dt = \int_0^2 x dx = \frac{1}{2}[x^2]_0^2 = \frac{1}{2}[4 - 0] = 2.$

Here we made the substitutions:

$\displaystyle \int_0^{\pi} \cos(t)e^{\sin(t)} dt = \int_1^{-1} e^x dx = [e^x]_1^{-1} = e^{-1} - e.$

Here we made the substitutions:
next up previous
Next: Substitutions to eliminate the Up: Internet Calculus II: Substitution Previous: Examples
George A. J. Sparling 2002-02-03