Answers to exercises

Exercise 1

Since \(x_0, x_1, \dots, x_j\) are all separated by distance \(\frac{1}{n} (b-a) = \Delta x\), we have

\begin{align*} x_j &= (x_j - x_{j-1}) + (x_{j-1} - x_{j-2}) + \dots + (x_1 - x_0) + x_0 \\ &= j \, \Delta x + x_0 \\ &= a + j \, \Delta x. \end{align*}

In the last equality we used the fact that \(x_0 = a\), by definition.

Exercise 2

The two subintervals are \([0,2]\) and \([2,4]\), each of width \(\Delta x = 2\), with respective midpoints 1 and 3. So the area estimate is

\[ \big[ f(1) + f(3) \big] \, \Delta x = ( 2 + 12 ) \cdot 2 = 28. \]

Exercise 3

Over each subinterval \([x_{j-1},x_j]\), the three estimates give rectangles of the same width but different heights. For the left-endpoint, midpoint and right-endpoint estimates, respectively, the heights are \(f(x_{j-1})\), \(f \big( \frac{1}{2} (x_{j-1} + x_j) \big)\) and \(f(x_j)\). Now by definition \(x_{j-1} < x_j\) and obviously the midpoint \(\frac{1}{2} (x_{j-1} + x_j)\) lies between them, so

\[ x_{j-1} < \frac{x_{j-1} + x_j}{2} < x_j. \]

As \(f\) is an increasing function, we have

\[ f(x_{j-1}) \leq f \Big( \frac{x_{j-1}+x_j}{2} \Big) \leq f(x_j). \]

It follows that the rectangles for the left-endpoint estimate are shorter than the rectangles for the midpoint estimate, which are shorter again than those for the right-endpoint estimate. This gives the desired inequalities.

When \(f\) is decreasing, we have the opposite inequalities

\[ f(x_{j-1}) \geq f \Big( \frac{x_{j-1}+x_j}{2} \Big) \geq f(x_j), \]

which imply

\[ \text{left-endpoint estimate } \geq \text{ midpoint estimate } \geq \text{ right-endpoint estimate}. \]

Exercise 4

With the interval \([a,b]\) divided into \(n\) subintervals \([x_{j-1}, x_j]\), for \(j=1, \dots, n\), and \(\Delta x = \frac{1}{n} (b-a)\), as usual, we have

\begin{align*} \text{left-endpoint estimate} &= \big[ f(x_0) + f(x_1) + \dots + f(x_{n-1}) \big] \, \Delta x,\\ \text{right-endpoint estimate} &= \big[ f(x_1) + f(x_2) + \dots + f(x_n) \big] \, \Delta x,\\ \text{average} &= \frac{1}{2} \Big( \big[ f(x_0) + \dots + f(x_{n-1}) \big] \, \Delta x + \big[ f(x_1) + \dots + f(x_n) \big] \, \Delta x \Big) \\ &= \frac{1}{2} \Big( f(x_0) + 2 f(x_1) + \dots + 2 f(x_{n-1}) + f(x_n) \Big) \, \Delta x \\ &= \Big[ \frac{1}{2} f(x_0) + f(x_1) + \dots + f(x_{n-1}) + \frac{1}{2} f(x_n) \Big] \, \Delta x \\ &= \text{trapezoidal estimate.} \end{align*}

Exercise 5

Divide \([0,1]\) into \(n\) subintervals \([x_0, x_1], \dots, [x_{n-1}, x_n]\), so \(x_j = \frac{j}{n}\) and \(\Delta x = \frac{1}{n}\). The right-endpoint estimate for the area is then

\[ \sum_{j=1}^n f(x_j) \; \Delta x = \sum_{j=1}^n f \Big( \frac{j}{n} \Big) \; \frac{1}{n} = \sum_{j=1}^n \Big( \frac{j}{n} \Big) \; \frac{1}{n} = \frac{1}{n^2} \sum_{j=1}^n j. \]

Using the formula \(\sum_{j=1}^n j = \frac{1}{2} n(n+1)\) gives the right-endpoint estimate as

\[ \frac{n(n+1)}{2n^2} = \frac{n^2 + n}{2n^2} = \frac{1}{2} \Big(1 + \frac{1}{n}\Big). \]

Taking the limit as \(n \to \infty\), the exact area under the curve is

\[ \lim_{n \to \infty} \frac{1}{2} \Big(1 + \frac{1}{n}\Big) = \frac{1}{2}. \]

The area under the curve is just a right-angled triangle with height 1 and base 1, so we can confirm that its area is \(\frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}\).

Exercise 6

\(f'(x) = x^n\)
\(f'(x) = (ax+b)^n\).

Exercise 7

We can rewrite this as \(\int (-2x+3)^{-\frac{1}{2}} \; dx\), so the integrand is of the form \((ax+b)^n\). We obtain \(-(-2x+3)^{\frac{1}{2}} + c = - \sqrt{3-2x} + c\).
Rewriting as \(\int (-2x+3)^{-2} \; dx\), the integrand is again of the form \((ax+b)^n\),and we obtain \[ \frac{1}{2} (-2x+3)^{-1} + c = \frac{1}{2(3-2x)} + c. \]

Exercise 8

\(\displaystyle \int \big(3x^2 + x^{\frac{1}{3}} \big) \; dx = x^3 + \frac{3}{4} x^{\frac{4}{3}} + c\)

Exercise 9

Let \(F(x)\) be an antiderivative of \(f(x)\) and let \(G(x)\) be an antiderivative of \(g(x)\), so \(\int f(x) \; dx = F(x) + c_1\) and \(\int g(x) \; dx = G(x) + c_2\), where \(c_1, c_2\) are constants. Then the derivative of \((F \pm G)(x)\) is \(F'(x) \pm G'(x) = f(x) \pm g(x)\), hence \(F \pm G\) is an antiderivative of \(f \pm g\), and \(\int \big(f(x) \pm g(x)\big) \; dx = F(x) \pm G(x) + c\), where \(c\) is a constant. Noting that the sum of two constants is a constant gives the desired equality.
Let \(F(x)\) be an antiderivative of \(f(x)\), so \(\int f(x) \; dx = F(x) + c\), where \(c\) is a constant. Then the derivative of \((kF)(x)\) is \(kF'(x) = kf(x)\), hence \(kF\) is an antiderivative of \(kf\) and \(\int k f(x) \; dx = k F(x) + C\), where \(C\) is a constant. As a constant times \(k\) is another constant, we have the desired equality.

Exercise 10

\(\displaystyle \int_0^8 (3x^2 + \sqrt[3]{x}) \; dx = \Bigg[ x^3 + \frac{3}{4} x^{\frac{4}{3}} \Bigg]_0^8 = \Big( 512 + \frac{3}{4} \cdot 16 \Big) - \big(0 + 0\big) = 512 + 12 = 524\)

Exercise 11

The graph \(y=x+1\) crosses the \(x\)-axis at \(x=-1\), so the desired area is

\begin{align*} \int_{-1}^2 (x+1) \; dx \, - \int_{-2}^{-1} (x+1) \; dx &= \Bigg[ \frac{1}{2} x^2 + x \Bigg]_{-1}^2 - \Bigg[ \frac{1}{2} x^2 + x \Bigg]_{-2}^{-1} \\ &= \Bigg( \big( 2 + 2 \big) - \Big( \frac{1}{2} - 1 \Big) \Bigg) - \Bigg( \Big( \frac{1}{2} - 1 \Big) - \big( 2 - 2 \big) \Bigg) \\ &= \Bigg( 4 - \Big( {-} \frac{1}{2} \Big) \Bigg) - \Big( {-} \frac{1}{2} - 0 \Big) = \frac{9}{2} + \frac{1}{2} = 5. \end{align*}

Alternatively, we can compute the areas of the triangles directly. The left triangle has height 1 and base 1, and so area \(\frac{1}{2}\). The right triangle has height 3 and base 3, and so area \(\frac{9}{2}\). The total area is 5.

Exercise 12

\(\displaystyle \int_{-2}^2 (x^3 - x) \; dx = \Bigg[ \frac{1}{4} x^4 - \frac{1}{2} x^2 \Bigg]_{-2}^2 = ( 4 - 2 ) - ( 4 - 2 ) = 0\)
Factorising \(x^3 - x = (x+1)x(x-1)\), we see the graph has intercepts at \(x=-1,0,1\). It is above the \(x\)-axis for \(-1 < x <0\) and \(x > 1\), and below the \(x\)-axis for \(x < -1\) and \(0 < x < 1\). We compute four separate integrals: \begin{align*} \int_{-2}^{-1} (x^3 - x) \; dx &= \Bigg[ \frac{1}{4} x^4 - \frac{1}{2} x^2 \Bigg]_{-2}^{-1} = \Big( \frac{1}{4} - \frac{1}{2} \Big) - \big( 4 - 2 \big) = - \frac{1}{4} - 2 = - \frac{9}{4},\\ \int_{-1}^0 (x^3 - x) \; dx &= \Bigg[ \frac{1}{4} x^4 - \frac{1}{2} x^2 \Bigg]_{-1}^0 = \big( 0 \big) - \Big( \frac{1}{4} - \frac{1}{2} \Big) = \frac{1}{4}, \\ \int_0^1 (x^3 - x) \; dx &= \Bigg[ \frac{1}{4} x^4 - \frac{1}{2} x^2 \Bigg]_0^1 = \Big( \frac{1}{4} - \frac{1}{2} \Big) - \big( 0 \big) = - \frac{1}{4}, \\ \int_1^2 (x^3 - x) \; dx &= \Bigg[ \frac{1}{4} x^4 - \frac{1}{2} x^2 \Bigg]_1^2 = \big( 4 - 2 \big) - \Big( \frac{1}{4} - \frac{1}{2} \Big) = \frac{9}{4}. \end{align*} Thus the total area is \[ \frac{9}{4} + \frac{1}{4} + \frac{1}{4} + \frac{9}{4} = 5. \]

Exercise 13

The equation \(\int_a^b k \, f(x) \; dx = k \int_a^b f(x) \; dx\) has the geometric interpretation that the signed area under the graph of \(y=kf(x)\) between \(x=a\) and \(x=b\) is \(k\) times the area under the graph of \(y=f(x)\). Indeed, the graph of \(y=k f(x)\) is obtained from the graph of \(y=f(x)\) by a dilation of factor \(k\) from the \(x\)-axis. Algebraically, letting \(F\) be an antiderivative of \(f\), we have \(\int_a^b kf(x) \; dx = \big[ k F(x) \big]_a^b = kF(b) - kF(a) = k \big[ F(x) \big]_a^b = k \int_a^b f(x) \; dx\).

Exercise 14

If \(a < b\), then the equation follows immediately from the fundamental theorem of calculus. If \(a>b\), then \(\int_a^b f(x) \; dx = - \int_b^a f(x) \; dx = - \big(F(a) - F(b)\big) = F(b) - F(a)\), as desired.

Exercise 15

As a reality check: for \(x\) between 0 and 5, the integrand is negative; and \(x\) is going backwards from 5 to 0; so the answer should be positive.

\begin{align*} \int_5^0 (-2x-3) \; dx = \big[ {-}x^2 - 3x \big]_5^0 = ( 0 - 0 ) - ( - 25 - 15 ) = 0 - (-40) = 40. \end{align*}

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