The integral of the expression $x^n$ where $n \neq -1$ is:

$\int x^ndx = \frac{x^{n+1}}{n+1} + C$

The integral of $x^{-1}$ is:

$\int \frac{1}{x}\,dx = \ln \left| x \right| + C$

Note the absolute value inside the natural log.

The integral of the exponential function $e^x$ is:

$\int e^x \, dx = e^x + C$

## Find a: $\displaystyle \int_0^{\ln a} e^x \, dx = 100$

Since $\int e^x \, dx = e^x + C$, we see that:

$\begin{aligned} \int_0^{\ln a} e^x \, dx &= 100 \\ \left[ e^x \right]_0^{\ln a} &= 100 \\ e^{\ln a}-e^0 &= 100 \\ a &= 101 _\square \end{aligned}$

## Evaluate: $\displaystyle \int_0^1 7x^6 + 42 + \frac{2}{x^3} \, dx$.

We can rewrite this problem so that all terms are in the form $x^n$:

$\displaystyle \int_0^1 7x^6 + 42x^0 + 2x^{-3} \, dx$

Now, let's integrate, using the rules above:

$\begin{aligned} \int_0^1 7x^6 + 42 + 2x^{-3} \, dx &= \int_0^1 7x^6 dx + \int_0^1 42 x^0 \, dx + \int_0^1 2x^{-3}\, dx \\ &= \left[ 7\left(\frac{x^{6+1}}{6+1} \right ) + 42x^1 + 2 \left( \frac{x^{-3+1}}{-3+1} \right ) \right]_0^1 \\ &=\left. x^7 + 42x -x^{-2} \right|_0^1 \\ &= \left(1^7 +42(1)-1\right)-\left(0\right) \\ &= 42 \, _\square \end{aligned}$

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