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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

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

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Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.

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