7.1: Power, exponential, trig, and logarithm rules
We’ve already seen the inverse power rule, but here it is again:
\(\boxed{\int x^mdx = \frac{x^{m + 1}}{m+1} + C}\)
Note that this only works if \(m \neq -1\). 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.
\(\boxed{\int e^xdx = e^x + C}\)
\(\boxed{\int \frac{1}{x}dx = \ln(x) + C}\)
\(\boxed{\int \sin(x)dx = -\cos(x) + C}\)
\(\boxed{\int \cos(x)dx = \sin(x) + C}\)
On to the examples. Recall that \(x^{-m} = \frac{1}{x^m}\) and .
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Find .
Negative and fractional powers work the same way as the positive powers we’ve been working with. We see
At this point, let’s just type it into a calculator to get an approximate answer. We see that the integral is about \(\boxed{23.36}\).
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Find \(\int_1^2 \frac{1}{x^4}dx\).
This problem becomes an inverse power rule problem if we notice that \(\frac{1}{x^4} = x^{-4}\). We see
\[\begin{align*} \int_1^2 \frac{1}{x^4}dx & = \int_1^2 x^{-4}dx \\ & = \frac{x^{-5}}{-5} \Big|_1^2 \\ & = -\frac{1}{5x^5} \Big|_1^2 \\ & = \left( -\frac{1}{5(2)^5} \right) - \left( -\frac{1}{5(1)^5} \right) \\ & = -\frac{1}{160} + \frac{1}{5} \\ & = -\frac{1}{160} + \frac{32}{160} \\ & = \boxed{\frac{31}{160}}. \end{align*}\]
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Find \(\int_{25}^{100} \sqrt{x}dx\).
If we remember that , this is a power rule problem.
Now for logarithmic and exponential functions.
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Find \(\int_1^5 \frac{2}{x}dx\).
For this problem, you may be tempted to write this as \(2x^{-1}\) 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.
*** QuickLaTeX cannot compile formula: \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*} *** Error message: Error: get_image_size(): -1
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Find \(\int_{-2}^2 7 e^{x}dx\).
\(e^x\) 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: \(\approx \boxed{50.78}\).