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1.

Approximate the area of the region under the curve of

\begin{equation*} y = \frac{4 - x}{1 + x^2} \end{equation*}

by calculating the area the area of the shaded rectangles shown in the diagram below. Which approximation method do these rectangles represent?

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Figure A.1.1. The Graph of \(y = \dfrac{4 - x}{1 + x^2}\) with the region under the curve on \([-3,3]\)
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Solution.

Let \(f(x) = \frac{4-x}{1+x^2}\text{.}\) Each rectangle in the diagram has a width of 1 and the height is taken as the function value at the midpoint of each subinterval. Thus, the area of the shaded rectangles can be approximated by

\begin{align*} \text{Area} \amp\approx 1\cdot\left( f\left(\frac{-3-2}{2}\right) + f\left(\frac{-2-1}{2}\right) + f\left(\frac{-1+0}{2}\right) + f\left(\frac{0+1}{2}\right) + f\left(\frac{1+2}{2}\right) + f\left(\frac{2+3}{2}\right)\right) \\ \amp= 1\cdot\left( f(-2.5) + f(-1.5) + f(-0.5) + f(0.5) + f(1.5) + f(2.5)\right)\\ \amp= 1\cdot\left( \frac{26}{29} + \frac{22}{13} + \frac{18}{5} + \frac{14}{5} + \frac{10}{13} + \frac{6}{29} \right) \\ \amp= \frac{18784}{1885} \approx 9.965 \end{align*}

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Note A.1.2.

Graphically, it is clear to see that the heights of each rectangle, using the midpoint rule, are the function values of midpoints of each subinterval, not the midpoint of function values. In fact, if you took the midpoints of the function values, i.e., you did something like

\begin{align*} \text{Area} \approx \frac{f(-3) + f(-2)}{2} + \frac{f(-2) + f(-1)}{2} \amp+ \frac{f(-1) + f(0)}{2} \\ \amp+ \frac{f(0) + f(1)}{2} + \frac{f(1) + f(2)}{2} + \frac{f(2) + f(3)}{2} \end{align*}

You actually used the trapezoidal rule instead of the midpoint rule. We will learn this new rule in Chapter 7.

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2.

Justify grapically, with pretty pictures and explanation that

If \(f\) is a continuous odd function, then \(\displaystyle \int_{-a}^a f(x)\, dx = 0\)
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If \(g\) is a continuous even function, then \(\displaystyle \int_{-a}^a g(x)\, dx = 2\int_0^a g(x)\, dx\)
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Hint.

Recall that

\(f\) is an odd function if its graph is symmetric across the origin.
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\(g\) is an even function if its graph is symmetric about the \(\boldsymbol{y}\)-axis.
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Solution.

Let’s consider an odd function \(y = f(x)\) first. Graphically, we define an odd function as one symmetric about the origin. That is, you can obtain the same figure by rotating the graph by \(180^\circ\text{.}\) The graph of an odd function looks like the one in the figure below.

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Observe that the area of the regions above and below the \(x\)-axis are identical because of symmetry. That is, the green region and the orange region has the same area. Since we are considering the signed area, then the green region will a the positive area and the orange region will have a negative area. So they canceled out when added. That is,

\begin{equation*} \int_{-a}^a f(x)\, dx = \int_{-a}^0 f(x)\, dx + \int_0^a f(x)\, dx = 0 \end{equation*}

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Now let’s consider an even function \(y = g(x)\text{.}\) Graphically, we define an even function as one symmetric across the \(y\)-axis. That is, you can obtain the same figure by reflecting the graph across the \(y\)-axis. The graph of an even function looks like the one in the figure below.

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Observe that the area of the regions on two sides of the \(y\)-axis are identical because of symmetry. That is, the green region and the orange region has the same area. Since we are considering the signed area, then the green region will have a positive area and the orange region will also have a positive area. So they double out when added. That is,

\begin{equation*} \int_{-a}^a g(x)\, dx = \int_{-a}^0 g(x)\, dx + \int_0^a g(x)\, dx = 2\int_0^a g(x)\, dx \end{equation*}

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Note A.1.3.

One of the goals for this problem is to potentially give you a shortcut on taking integrals. If you see that the limits of integration are opposite numbers, then you may want to check the parity (even or odd) of the function.

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If the function is odd, then we can simply say that the integral equals to \(0\text{;}\) if the function is even, then we can replace the lower limit with \(0\) and multiply the integral by \(2\text{,}\) as plugging in \(0\) to a function is always easier than plugging in an negative number.

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3.

Find constants \(c_1\) and \(c_2\) such that

\begin{equation*} F(x) = c_1\sin(3x) + c_2x\cos(3x) \end{equation*}

is an antiderivative of \(f(x) = 2x\sin(3x)\text{.}\)

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Hint.

Recall that \(F\) is an antiderivative of \(f\) IF \(F'(x) = f(x)\text{.}\)

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Don’t try to evaluate \(\int f(x)\, dx\) as we need to build up more concepts to be able to evaluate this integral. Instead, take the derivative of \(F(x)\) and match it up with \(f(x)\text{.}\)

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We want to find \(c_1\) and \(c_2\) such that \(F'(x)\) and \(f(x)\) look exactly the same.

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Solution.

Using the definition of antiderivatives, we can set up an equation such that \(\int f(x) \, dx = F(x) + C\text{.}\) Yet, we can’t evaluate the integral \(\int 2x\sin(3x)\,dx\) so far, as we haven’t learned the technique to evaluate this integral yet (the technique is called Integration by parts, which is what Section 7.1 is about). However, we can rewrite the equation using derivatives. That is, \(F'(x) = f(x)\text{.}\) So let’s take the derivative of \(F(x)\text{.}\)

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\begin{align*} F'(x) \amp= \frac{d}{dx} \left( c_1\sin(3x) + c_2 x\cos(3x)\right)\\ \amp= c_1 \frac{d}{dx}\sin(3x) + c_2 \frac{d}{dx} (x\cos(3x)) \amp\amp \text{by the Linearity of Derivatives}\\ \amp= c_1\cdot 3\cos(3x) + c_2 (\cos(3x) - 3x\sin(3x)) \amp\amp \text{by the Chain Rule and the Product Rule}\\ \amp= 3c_1\cos(3x) + c_2 \cos(3x) - 3c_2 x\sin(3x)\\ \amp= (3c_1 + c_2) \cos(3x) - 3c_2 x\sin(3x) \end{align*}

We know that this result must equal \(f(x)\text{.}\) That is,

\begin{equation*} (3c_1 + c_2) \cos(3x) - 3c_2 x\sin(3x) = 2x\sin(3x) \end{equation*}

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Two functions are the same if they look exactly the same, including the coefficients. Both \(F'(x)\) and \(f(x)\) have a term of \(x\sin(3x)\text{,}\) so the coefficients of this term must be the same. That is,

\begin{equation*} -3c_2 = 2 \qquad\implies\qquad c_2 = -\frac{2}{3} \end{equation*}

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But there is no \(\cos(3x)\) in \(f(x)\text{.}\) That is, we need to find out a way to eliminate the term \((3c_1 + c_2)\cos(3x)\) in \(\frac{d}{dx}\left(F(x)\right)\text{.}\) The way to do so is to set the coefficient of this term to 0. That is,

\begin{equation*} 3c_1 + c_2 = 0 \qquad\implies\qquad c_1 = -\frac{c_2}{3} = -\frac{-2/3}{3} = \frac{2}{9} \end{equation*}

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Hence, the two constants that make \(F(x)\) an antiderivative of \(f(x)\) are \(c_1 = \frac{2}{9}\) and \(c_2 = -\frac{2}{3}\text{.}\)

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4.

Find \(f'\) and \(f\) using the following conditions:

\begin{equation*} f''(x) = x - \cos(x), \qquad f'(0) = 2, \qquad f(0) = -2 \end{equation*}

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Solution.

By FTC, we know that integration undoes differentiation. That is,

\begin{equation*} f'(x) = \int f''(x) \, dx \hspace{20px}\text{ and }\hspace{20px} f(x) = \int f'(x) \, dx \end{equation*}

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So let’s start on taking the integrals!

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\begin{align*} f'(x) = \int f''(x) \, dx \amp= \int \left( x - \cos(x) \right) \, dx\\ \amp= \int x \, dx - \int \cos(x)\, dx \amp\amp \text{by the Linearity of Integrals}\\ \amp= \frac{x^2}{2} - \sin(x) + C_1 \amp\amp \text{by the Power Rule and Trig Rule} \end{align*}

We can determine what \(f'(x)\) is specifically using the initial condition \(f'(0) = 2\text{.}\) That is,

\begin{equation*} 2 = f'(0) = \frac{0^2}{2} - \sin(0) + C_1 \hspace{20px}\implies\hspace{20px} C_1 = 2 \end{equation*}

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Hence, we know that \(f'(x) = \dfrac{x^2}{2} - \sin(x) + 2\text{.}\) We can take one more integral to obtain \(f(x)\text{.}\)

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\begin{align*} f(x) = \int f'(x) \, dx \amp= \int \left( \frac{x^2}{2} - \sin(x) + 2 \right) \, dx\\ \amp= \frac{1}{2} \int x^2 \, dx - \int \sin(x)\, dx + 2 \int \, dx \amp\amp \text{by the Linearity of Integrals}\\ \amp= \frac{x^3}{6} + \cos(x) + 2x + C_2 \amp\amp \text{by the Power Rule and Trig Rule} \end{align*}

We can determine what \(f(x)\) is specifically using the initial condition \(f(0) = -2\text{.}\) That is,

\begin{equation*} -2 = f(0) = \frac{0^3}{6} + \cos(0) + 2(0) + C_2 \hspace{20px}\implies\hspace{20px} C_2 = -3 \end{equation*}

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Hence, we know that \(f(t) = \dfrac{x^3}{6} + \cos(x) + 2x - 3\text{.}\)

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Worksheet 

1.

Note A.1.2.

2.

Note A.1.3.

3.

4.

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