A Modeling Approach to Calculus: MA2C

Let us consider the two steps separately. Then we will see how to represent this more compactly using evaluation notation.

First, we need an antiderivative, computed as an indefinite integral

This tells us that \(F(x) = \frac{1}{3}x^3\) is an antiderivative, as is \(F(x) = \frac{1}{3}x^3+4\) (or any other constant). We'll use the first, since it is simpler; we only need one antiderivative, not all of them.

Second, the Fundamental Theorem of Calculus allows us to evaluate the definite integral as the change in \(F(x)\text{.}\)

This can be written more compactly by writing the formula of the antiderivative inside the evaluation notation while simultaneously indicating the use of the Fundamental Theorem of Calculus:

Notice that we did not need the constant of integration because the Fundamental Theorem of Calculus only requires one antiderivative. We generally choose the most convenient one with a zero constant.

The integrand has a product of \(x\) with \(u^4\) where \(u=3x^2+1\text{.}\) For the chain rule to have been the source of this product, we would need \(\frac{du}{dx} = 6x\) rather than \(x\text{.}\)

The integrand is of the form \(u^{-1}\) where \(u=2x-1\text{.}\) The derivative of \(u\) is \(u' = 2\text{.}\) In order to antidifferentiate the chain rule, we rewrite

To see if the antiderivative can be used in a definite integral, we need to see where \(f(x) = \frac{1}{2x-1}\) is continuous. A discontinuity occurs at \(x=\frac{1}{2}\text{.}\) Consequently, a definite integral using the antiderivative can only be used for intervals that do not include \(\frac{1}{2}\text{.}\) Thus, \(\displaystyle \int_0^2 \frac{1}{2x-1}\,dx\) can not be computed. On the other hand,

When solving problems involving definite integrals, it is often helpful to explicitly remind yourself of the concept of accumulation functions and the fundamental theorem of calculus's conclusion.

1. Define \(\displaystyle A(x) = \int_1^x e^{-z^2} \, dz\text{,}\) which is the accumulation function with integrand \(f(z) = e^{-z^2}\text{.}\) The Fundamental Theorem of Calculus tells us that \(A'(x) = f(x) = e^{-x^2}\text{.}\) The following work would communicate these results:

   \begin{align*} A(x) &= \int_1^x e^{-z^2} \, dz\\ A'(x) &\overset{{\mathrm{FTC}}}{=} e^{-x^2}\\ \frac{d}{dx} \int_1^x e^{-z^2} \, dz &= A'(x) = e^{-x^2} \end{align*}

2. To compute \(\displaystyle \frac{d}{dx} \int_x^{x^2} e^{-z^2} \, dz\text{,}\) we first need the accumulation function \(\displaystyle A(x) = \int_1^x e^{-z^2} \, dz\) with rate \(f(x)=e^{-x^2}\text{.}\) The integral that defines our function involves a composition by the splitting property,

   \begin{equation*} \int_x^{x^2} e^{-z^2} \, dz = A(x^2) - A(x). \end{equation*}

   When we differentiate this, we must use the chain rule knowing

   \begin{equation*} \frac{d}{dx}\Big[ A(u) \Big] = A'(u) \frac{du}{dx} \overset{\mathrm{FTC}}= f(u) \frac{du}{dx}. \end{equation*}

   The following work would communicate these results:

   \begin{align*} A(x) &= \int_1^x e^{-z^2} \, dz\\ \frac{d}{dx} \int_x^{x^2} e^{-z^2} \, dz &= \frac{d}{dx}\Big[A(x^2)-A(x)\Big]\\ &\overset{\mathrm{FTC}}= f(x^2) \cdot \frac{d}{dx}[x^2] - f(x)\\ &= e^{-(x^2)^2} \cdot 2x - e^{-x^2} \\ &= 2xe^{-x^4} - e^{-x^2} \end{align*}

3. To compute \(\displaystyle \frac{d}{dx}\left[ \int_0^{\sqrt{x}} \frac{1}{\sqrt{z^4+1}}\,dz \right]\text{,}\) we will need to define an accumulation function and then apply the Fundamental Theorem of Calculus to find the derivative required.

   \begin{align*} A(x) &= \int_0^x \frac{1}{\sqrt{z^4+1}} \, dz\\ A'(x)&\overset{{\mathrm{FTC}}}{=} \frac{1}{\sqrt{x^4+1}}\\ \frac{d}{dx}\left[ \int_0^{\sqrt{x}} \frac{1}{\sqrt{z^4+1}}\,dz \right] &= \frac{d}{dx}[A(\sqrt{x})]\\ &= A'(\sqrt{x}) \frac{d}{dx}[\sqrt{x}] \\ &= \frac{1}{\sqrt{(\sqrt{x})^4+1}} \cdot (\frac{1}{2}x^{-1/2}) \\ &= \frac{1}{\sqrt{x^2+1}} \cdot \frac{1}{2 \sqrt{x}} \\ &= \frac{1}{2 \sqrt{x(x^2+1)}} \end{align*}