Chapter 8: Review

The chapter problem, solved

The curved window

The window is the region enclosed between the curve $y = x^2$ and the line $y = x + 2$ .

Step 1. Find the crossing points. Set the two expressions equal:

x^2 = x + 2, \qquad x^2 - x - 2 = 0, \qquad (x - 2)(x + 1) = 0.

The edges cross at $x = -1$ and $x = 2$ .

Step 2. Decide which edge is upper. Test $x = 0$ : the line gives $2$ and the curve gives $0$ . The line is the upper edge between the crossings.

Step 3. Integrate the height. The height of the glass at each $x$ is $(x + 2) - x^2$ :

A = \int_{-1}^{2} \left((x + 2) - x^2\right)dx = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}.

At the upper limit:

\frac{4}{2} + 4 - \frac{8}{3} = 2 + 4 - \frac{8}{3} = \frac{10}{3}.

At the lower limit:

\frac{1}{2} - 2 + \frac{1}{3} = \frac{5}{6} - 2 = -\frac{7}{6}.

Subtract:

A = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}.

The glass has area $\dfrac{9}{2}$ square units, or $4.5$ square units.

The cooling drink

The coffee starts at $90$ °C in a room at $20$ °C, and after $20$ minutes it is $55$ °C. The temperature $T$ satisfies

\frac{dT}{dt} = -k(T - 20).

Step 1. Separate and integrate.

\int \frac{1}{T - 20}dT = \int -kdt, \qquad \ln|T - 20| = -kt + C.

Exponentiate and write $A = e^C$ :

T = 20 + Ae^{-kt}.

Step 2. Use the initial temperature. At $t = 0$ , $T = 90$ , so $90 = 20 + A$ and $A = 70$ :

T = 20 + 70e^{-kt}.

Step 3. Use the second temperature to find $k$ . At $t = 20$ , $T = 55$ :

55 = 20 + 70e^{-20k}, \qquad e^{-20k} = \frac{35}{70} = \frac{1}{2}, \qquad k = \frac{\ln 2}{20}.

T = 20 + 70e^{-(\ln 2)t/20}.

Step 4. Find the temperature after 40 minutes. The exponent is $-40 \times \dfrac{\ln 2}{20} = -2\ln 2$ , so

T = 20 + 70e^{-2\ln 2} = 20 + 70 \times \frac{1}{4} = 20 + 17.5 = 37.5 \text{°C}.

Step 5. Describe the long run. As $t \to \infty$ the exponential term tends to $0$ , so the temperature approaches $20$ °C, the room temperature. The model predicts that the coffee never quite reaches room temperature in finite time.

Limitations. The model assumes that $k$ stays constant, that the room stays at $20$ °C, and that the coffee is well mixed at every moment. It also assumes a single smooth temperature, but a real cup has warmer and cooler regions. Once the coffee is close to room temperature the model adds little, because the change is smaller than the natural variation in the measurement.

Mixed practice — choose the method first

For each question, say which method you will use before you calculate: a standard integral, a substitution, integration by parts or partial fractions. The items mix this chapter with earlier chapters.

Find $\displaystyle\int \left(6x^2 - 4x + 1\right)dx$ .
Evaluate $\displaystyle\int_0^1 \left(3x^2 + 2x\right)dx$ .
Evaluate $\displaystyle\int_1^{e} \frac{1}{x}\,dx$ .
Find the area under the curve $y = 9 - x^2$ between $x = 0$ and $x = 3$ .
Find the area enclosed between the curve $y = x^2$ and the line $y = 2x + 3$ .
Find $\displaystyle\lim_{n \to \infty} \frac{3n + 1}{n}$ .
Find $\displaystyle\int xe^{x^2}\,dx$ .
Find $\displaystyle\int x\cos 2x\,dx$ .
Find $\displaystyle\int \frac{1}{(x + 2)(x - 1)}\,dx$ .
Solve $\dfrac{dy}{dx} = 2y$ for which $y = 5$ when $x = 0$ .
A cooling model gives $T = 25 + 55e^{-0.2t}$ . Find the initial temperature and the limiting temperature.
Differentiate $y = x^2\ln x$ .
Solve $3^x = 20$ , giving an exact answer for $x$ .
Evaluate $\displaystyle\int_0^{\pi/2} \sin 2x\,dx$ .

Answers and concise solutions

1. The integral

Answer: Integrate term by term: $6 \times \dfrac{x^3}{3} - 4 \times \dfrac{x^2}{2} + x + C = 2x^3 - 2x^2 + x + C$ .

2. The value of the integral

Answer: $\Big[x^3 + x^2\Big]_0^1 = (1 + 1) - 0 = 2$ .

3. The value of the integral

Answer: $\Big[\ln x\Big]_1^{e} = \ln e - \ln 1 = 1 - 0 = 1$ .

4. The area

Answer: The curve is above the axis, so the area is $\displaystyle\int_0^3 (9 - x^2)\,dx = \left[9x - \dfrac{x^3}{3}\right]_0^3 = 27 - 9 = 18$ square units.

5. The area

Answer: Set $x^2 = 2x + 3$ : $x^2 - 2x - 3 = 0$ , so $(x - 3)(x + 1) = 0$ and $x = -1$ or $x = 3$ . The line is above the curve, so the area is $\displaystyle\int_{-1}^{3}(2x + 3 - x^2)\,dx = \dfrac{32}{3}$ square units.

6. The limit

Answer: Divide top and bottom by $n$ : $\dfrac{3 + 1/n}{1} \to 3$ as $n \to \infty$ .

7. The integral

Answer: Let $u = x^2$ , so $du = 2x\,dx$ . Then $\dfrac{1}{2}\displaystyle\int e^u\,du = \dfrac{1}{2}e^{x^2} + C$ . A substitution, because the derivative of the inside appears.

8. The integral

Answer: By parts with $u = x$ and $v = \dfrac{1}{2}\sin 2x$ : $\dfrac{x}{2}\sin 2x - \displaystyle\int \dfrac{1}{2}\sin 2x\,dx = \dfrac{x}{2}\sin 2x + \dfrac{1}{4}\cos 2x + C$ . Integration by parts, because the integrand is a product.

9. The integral

Answer: The split is $-\dfrac{1}{3(x + 2)} + \dfrac{1}{3(x - 1)}$ , so the integral is $-\dfrac{1}{3}\ln|x + 2| + \dfrac{1}{3}\ln|x - 1| + C$ . Partial fractions, because the denominator is a product of linear factors.

10. The particular solution

Answer: Separate to $\ln|y| = 2x + C$ , so $y = Ae^{2x}$ . The condition $y(0) = 5$ gives $A = 5$ , so $y = 5e^{2x}$ . A separable differential equation.

11. The initial and limiting temperatures

Answer: At $t = 0$ , $T = 25 + 55 = 80$ °C. As $t \to \infty$ the exponential term tends to $0$ , so the limiting temperature is $25$ °C, the surrounding temperature.

12. The derivative

Answer: A product, so use the product rule: $2x\ln x + x^2 \times \dfrac{1}{x} = 2x\ln x + x$ . This comes from Chapter 7.

13. The value of $x$

Answer: Take logarithms of both sides: $x\ln 3 = \ln 20$ , so $x = \dfrac{\ln 20}{\ln 3} \approx 2.73$ . This uses the logarithms from Chapter 6.

14. The value of the integral

Answer: $\displaystyle\int_0^{\pi/2} \sin 2x\,dx = \left[-\dfrac{1}{2}\cos 2x\right]_0^{\pi/2} = \dfrac{1}{2} - \left(-\dfrac{1}{2}\right) = 1$ .

What you can now do

use the Fundamental Theorem of Calculus to evaluate a definite integral;
find an antiderivative and use the constant of integration;
integrate powers, exponentials, trigonometric functions and $\dfrac{1}{x}$ ;
find the area under a curve, including where the curve crosses the axis;
find the area between two curves, splitting where they cross;
explain a definite integral as the limit of a sum;
integrate by substitution and by parts;
integrate a rational function using partial fractions;
solve a separable first-order differential equation;
interpret a differential-equation solution and name its limitations.

Your relevance reflection

Return to the prompt in the chapter opener: a quantity that grows or shrinks. Write the quantity again. Now add three things you could not have said before this chapter: how the total is built from many small pieces, how to find the total from the rate, and how to describe what happens in the long run. Finish with one sentence about a rate you would now measure, and the total you would calculate from it.