Mathematics for Management -- Supplementary Electronic Materials

Quiz: The Area Between Curves

After having worked through the electronic materials (set of notes and video clips), please self-evaluate your learning progress by completing the following quiz. Therefore, please check the correct answer of each of the given questions (single choice) and press the submit button after you completed all 10 questions. (Sometimes the page needs a reload such that the grading script starts to work.)

Your selections will persist so that you can review your answers in case you did not achieve the complete grade. At the bottom of the page you can also find the solution key.

1.The area of the region between the graph of \( f(x) = 4x^3 + 2 \) and the \( x \)-axis from \(x = 1\) to \(x = 2\) is .

\( 36 \) \( 23 \) \( 20 \) \( 17 \)

2.The area of the region enclosed by the graph of \( f(x) = x^2 + 2 \) and the line \( g(x) = 11 \) is

\( 18 \) \( 30 \) \( 36 \) \( 72 \)

3. The area of the region enclosed by the graphs of \( f(x) = x - 2x^2 \) and \( g(x) = -5x \) is

\( \frac{7}{3} \) \( \frac{16}{3} \) \( 9 \) \( 36 \)

4. The area of the region enclosed by the graphs of \( f(x) = -x^2 + x + 6 \) and \( y = 4 \) is

\( \frac{3}{2} \) \( \frac{31}{6} \) \( \frac{9}{2} \) \( \frac{33}{2} \)

5. The area of the region bounded by the curve \( y = {\rm{e}}^{2x} \), the \( x \)-axis, the \( y \)-axis, and the line \( x = 2 \) is

\( \frac{1}{2} {\rm{e}}^4 - {\rm{e}} \) \( \frac{1}{2} {\rm{e}}^4 - 1 \) \( \frac{1}{2} \left( {\rm{e}}^4 - 1 \right)\) \( 2 {\rm{e}}^4 - {\rm{e}} \)

6. The area of the region in the first quadrant that is enclosed by the graphs of \( f(x) = x^3 + 8 \) and \( g(x) = x + 8 \) is

\( \frac{1}{4} \) \( \frac{1}{2} \) \( \frac{3}{4} \) \( \frac{65}{4} \)

7.The area of the region enclosed by the curve \( y = \frac{1}{x-1} \), the \( x \)-axis, and the lines \( x = 3 \) and \( x = 4 \) is

\( \ln(\frac{2}{3}) \) \( \ln(\frac{3}{2}) \) \( \ln(\frac{4}{3}) \) \( \ln(\frac{3}{4})) \)

8. The integral \( \int^4_{-4} \sqrt{16 - x^2} \textrm{d} x \) gives the area of

a circle of radius \( 4 \) a semicircle of radius \( 4 \) a quadrant of a circle of radius \( 4 \) an ellipse whose semi-major axis is \( 4 \)

9. A region in the plane is bounded by the graph of \( f(x) = \frac{1}{x} \), the \( x \)-axis, the line \( x = m \), and the line \( x = 2m \), \( m > 0 \). The area of this region

is independent of \( m \) decreases as \( m \) increases decreases as \( m \) increases when \( m < \frac{1}{2} \) and increases as \( m \) increases when \( m > \frac{1}{2} \) increases as \( m \) increases when \( m < \frac{1}{2} \) and decreases as \( m \) increases when \( m > \frac{1}{2} \)

10. Let \( a > 0 \). The larger of the two areas into which the circle \( x^2 + y^2 = 16 a^2 \) is divided by the parabola \( y^2 = 6ax \) is

\( \frac{4}{3} a^2 \left( 8 \pi - \sqrt{3} \right) \) \( \frac{4}{3} a^2 \left( 4 \pi - \sqrt{3} \right) \) \( \frac{2}{3} a^2 \left( 4 \pi + \sqrt{3} \right) \) \( \frac{4}{3} a^2 \left( 4 \pi + \sqrt{3} \right) \)

Your grade is: __ of 10

In case you cannot find the correct answers even after reviewing the electronic materials again, please click on the "+" sign to read more.

Solution: 1d; 2c; 3c; 4d; 5c; 6a; 7b; 8b; 9a; 10a;

Here, a, b, c, d indicate the 1st, 2nd, 3rd, and 4th answer choice, respectively, for the numbered questions.