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 .

Power rule with fractional and negative powers

• 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 \(\boxed{23.36}\).

• 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

  \[\begin{align*} \int_1^2 \frac{1}{x^4}dx & = \int_1^2 x^{-4}dx \\ & = \frac{x^{-5}}{-5} \Big|_1^2 \\ & = -\frac{1}{5x^5} \Big|_1^2 \\ & = \left( -\frac{1}{5(2)^5} \right) - \left( -\frac{1}{5(1)^5} \right) \\ & = -\frac{1}{160} + \frac{1}{5} \\ & = -\frac{1}{160} + \frac{32}{160} \\ & = \boxed{\frac{31}{160}}. \end{align*}\]

• Find \(\int_{25}^{100} \sqrt{x}dx\).

  If we remember that , this is a power rule problem.

  

Now for logarithmic and exponential functions.

Logarithmic and Exponential functions and integration

• 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
  

• 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: \(\approx \boxed{50.78}\).