We’ve already seen the inverse power rule, but here it is again:
[latex]\boxed{\int x^mdx = \frac{x^{m + 1}}{m+1} + C}[/latex]
Note that this only works if [latex]m \neq -1[/latex]. However, we haven’t seen how this works with fractional and negative powers yet. We’ll do some examples of this.
We can also “undo” the derivatives for exponential, logarithmic functions, or trigonometric functions.
[latex]\boxed{\int e^xdx = e^x + C}[/latex]
[latex]\boxed{\int \frac{1}{x}dx = \ln(x) + C}[/latex]
[latex]\boxed{\int \sin(x)dx = -\cos(x) + C}[/latex]
[latex]\boxed{\int \cos(x)dx = \sin(x) + C}[/latex]
On to the examples. Recall that [latex]x^{-m} = \frac{1}{x^m}[/latex] and [latex]x^{1/m} = \sqrt[m]{x}[/latex].
Power rule with fractional and negative powers
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Find [latex]\int_4^9 x^{-3} + 2x^{1/2}dx[/latex].
Negative and fractional powers work the same way as the positive powers we’ve been working with. We see
\begin{align*}
\int_4^9 x^{-3} + 2x^{1/2}dx & = \int_4^9 x^{-3}dx + 2 \int_4^9 x^{1/2}dx \\
& = \left( \frac{x^{-2}}{-2} \right) + 2 \left( \frac{x^{3/2}}{3/2} \right) \Big|_4^9 \\
& = \frac{1}{-2x^2} + 2 \frac{2 x^{3/2}}{3} \Big|_4^9 \\
& = \frac{1}{-2x^2} + \frac{4 x^{3/2}}{3} \Big|_4^9 \\
& = \left( \frac{1}{-2(9)^2} + \frac{4 (9)^{3/2}}{3} \right) – \left( \frac{1}{-2(4)^2} + \frac{4 (4)^{3/2}}{3} \right) \\
& = \left( -\frac{1}{162} + \frac{108}{3} \right) – \left( -\frac{1}{32} + \frac{32}{3} \right)
\end{align*}
At this point, let’s just type it into a calculator to get an approximate answer. We see that the integral is about [latex]\boxed{23.36}[/latex]. -
Find [latex]\int_1^2 \frac{1}{x^4}dx[/latex].
This problem becomes an inverse power rule problem if we notice that [latex]\frac{1}{x^4} = x^{-4}[/latex]. We see
\begin{align*}
\int_1^2 \frac{1}{x^4}dx & = \int_1^2 x^{-4}dx \\
& = \frac{x^{-5}}{-5} \Big|_1^2 \\
& = -\frac{1}{5x^5} \Big|_1^2 \\
& = \left( -\frac{1}{5(2)^5} \right) – \left( -\frac{1}{5(1)^5} \right) \\
& = -\frac{1}{160} + \frac{1}{5} \\
& = -\frac{1}{160} + \frac{32}{160} \\
& = \boxed{\frac{31}{160}}.
\end{align*} -
Find [latex]\int_{25}^{100} \sqrt{x}dx[/latex].
If we remember that [latex]x^{1/2} = \sqrt{x}[/latex], this is a power rule problem.
\begin{align*}
\int_{25}^{100} \sqrt{x}dx & = \int_{25}^{100} x^{1/2}dx \\
& = \frac{x^{3/2}}{3/2} \Big|_{25}^{100} \\
& = \frac{2 x^{3/2}}{3} \Big|_{25}^{100} \\
& = \left( \frac{2 (100)^{3/2}}{3} \right) – \left( \frac{2 (25)^{3/2}}{3} \right) \\
& = \frac{2000}{3} – \frac{250}{3} \\
& = \boxed{\frac{1750}{3}}.
\end{align*}
Now for logarithmic and exponential functions.
Logarithmic and Exponential functions and integration
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Find [latex]\int_1^5 \frac{2}{x}dx[/latex].
For this problem, you may be tempted to write this as [latex]2x^{-1}[/latex] and use the inverse power rule. Good instincts, but in this case it won’t work (try it: it leads to division by zero!). So instead, let’s invert the natural logarithm derivative.
\begin{align*}
\int_1^5 \frac{2}{x}dx & = 2 \int_1^5 \frac{1}{x}dx \\
& = 2 \left( \ln(x) \right) \Big|_1^5 \\
& = (2 \ln(5)) – (2 \ln(1)) \\
& = 2 \ln(5) – 0 \\
& = 2 \ln(5) \approx \boxed{3.2}.
\end{align*} -
Find [latex]\int_{-2}^2 7 e^{x}dx[/latex].
[latex]e^x[/latex] is the best function for calculus. Doesn’t change with integration!
\begin{align*}
\int_{-2}^2 7 e^xdx & = 7 \int_{-2}^2 e^xdx \\
& = 7 (e^x) \Big|_{-2}^2 \\
& = (7 e^{2}) – (7 e^{-2})
\end{align*}
Not a lot to do to simplify this, but we can get a decimal approximation: [latex]\approx \boxed{50.78}[/latex].