# Calculus

## Integrals of Exponential Form

We know that the derivative of the exponential function $$de^u/dx=e^u\, du/dx$$. Reversing the differential, the integral of an exponential function is

 $\int e^udu=e^u+C$

Example 1 Integrate $\int 6x^2e^{x^3}dx$

Solution: let $$u=x^3$$,

\begin{align} du&= 3x^2dx \\ \frac{du}{3} &=x^2dx \end{align}

We can substitute into our original integral and get,

\begin{align} 2\int e^udu &=2e^u +C \\ &=2e^{x^3}+C \end{align}

Example 2: Integrate $\int \left(e^x-e^{-x}\right)^2dx$

Solution: We can simplify the expression in the integral to help us integrate.

\begin{align} \left(e^x-e^{-x}\right)^2 &= e^{2x}-2e^xe^{-x}+e^{-2x}\\ &=e^{2x}-2+e^{-2x} \end{align}

Thus,

$\int \left(e^{2x}-2+e^{-2x}\right) dx = \frac{e^{2x}}{2} -2x - \frac{e^{-2x}}{2} +C$

Example 3: Find the equation of the curve for which $$dy/dx=\sqrt{e^{x+3}}$$ if the curve passes through $$(1,0)$$.

Solution: First, we integrate to find the general equation form

\begin{align} y&=\int e^{\frac{x+3}{2}}dx \\ &= 2e^{\frac{x+3}{2}}+C\end{align}

Now we solve for C using the point $$(1,0)$$

\begin{align} 0&=2e^{\frac{1+3}{2}}+C \\ -2e^2&=C \end{align}

Thus, the equation of the curve is

$y=2e^{\frac{x+3}{2}}-2e^2$