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Suppose we have two continuous and differentiable functions, $f$ and $g$, with the properties that $f'(x) = g(x)$, $g^{3}(x)=(f(x)-1)^{2}, f(a)=1$, and $f(b)=2$.
	Compute the definite integral:
	\begin{align*}
		\int_{a}^{b} f^{2}(x)g^{4}(x)dx
	\end{align*}

\begin{align*}
	\int_{a}^{b} f^{2}(x)g^{4}(x)dx & = \int_{a}^{b}f^{2}(x)g^{3}(x)g(x)\ dx & \text{(rewrite)}  \\\\
	& = \int_{a}^{b}f^{2}(x)(f(x)-1)^{2}g(x)\ dx & \text{(substitute)}  \\\\
	& = \int_{a}^{b}f^{2}(x)[f^{2}(x)-2f(x)+1]g(x)\ dx & \text{(expand)} \\\\
	& = \int_{a}^{b}[f^{4}(x)-2f^{3}(x)+f^{2}(x)]g(x)\ dx & \text{(distribute)}\\\\
	& = \int_{a}^{b}f^{4}(x)g(x)-2f^{3}(x)g(x)+f^{2}(x)g(x)\ dx & \text{(distribute again)} \\\\
	& = \int_{a}^{b}f^{4}(x)g(x)\ dx - 2\int_{a}^{b}f^{3}(x)g(x)\ dx + \int_{a}^{b}f^{2}(x)g(x)\ dx & \text{(apply linearity)}	\\
\end{align*}

What is $g(x)$?  Right: $f'(x) = g(x)$:

\begin{align*}
	\int_{a}^{b} f^{2}(x)g^{4}(x)dx & = \int_{a}^{b}f^{4}(x)f'(x)\ dx - 2\int_{a}^{b}f^{3}(x)f'(x)\ dx + \int_{a}^{b}f^{2}(x)f'(x)\ dx & \text{(substitute)} \\\\	
\end{align*}

From Calc II, you should recognize that each of the terms on the right-hand-side of the equal sign has a function multiplied by the derivative of that function.  This is precisely the condition under which we would apply a u-substitution.  Watch what happens.  Let's proceed as normal with: 
\begin{align*}
	u & = f(x) \\
	du & = f'(x)dx \\
\end{align*}
Transforming our limits of integration, we have:
\begin{align*}
	u & = f(x) \\
	u & = f(a) = 1 & \text{(lower limit of integration)}\\
	u & = f(b) = 2 & \text{(upper limit of integration)}\\	
\end{align*}

Now, we have:
\begin{align*}
	\int_{a}^{b} f^{2}(x)g^{4}(x)dx & = \int_{a}^{b}f^{4}(x)f'(x)\ dx - 2\int_{a}^{b}f^{3}(x)f'(x)\ dx + \int_{a}^{b}f^{2}(x)f'(x)\ dx \\\\	
	& = \int_{1}^{2}u^{4}\ du - 2\int_{1}^{2}u^{3}\ du + \int_{1}^{2}u^{2}\ du\\
\end{align*}

Cleaned right up, didn't it?  Just finish it off!  :)

Thu, 07 Feb 2019 22:20 GMT