##### 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
1. $$f'(x) = x^n$$
2. $$f'(x) = (ax+b)^n$$.
##### Exercise 7
1. We can rewrite this as $$\int (-2x+3)^{-\frac{1}{2}} \; dx$$, so the integrand is of the form $$(ax+b)^n$$. We obtain $$-(-2x+3)^{\frac{1}{2}} + c = - \sqrt{3-2x} + c$$.
2. 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
1. 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.
2. 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
1. $$\displaystyle \int_{-2}^2 (x^3 - x) \; dx = \Bigg[ \frac{1}{4} x^4 - \frac{1}{2} x^2 \Bigg]_{-2}^2 = ( 4 - 2 ) - ( 4 - 2 ) = 0$$
2. 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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