UWorld AP® Calculus AB Practice Test Question Bank (QBank)
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Question
The graph of f ′, the derivative of the function f, is shown in the graph above. The function f is concave up on (2, 4).
| A. True | |
| B. False |
Explanation
The given graph shows the first derivative of the function f. Concavity is given by the second derivative, so the slope of f′ gives the concavity of f. The graph of f′ is increasing on (2, 3) but decreasing on (3, 4). Therefore, f is concave down on part of the interval (2, 4), and the given statement is false.
Question
What is the area of the region R bounded by the graphs of x = y2 − y + 1 and x = y + 1 ?
| A.
4
3
|
|
| B.
8
3
|
|
| C.
4
|
|
| D.
20
3
|
Explanation
The area between the given functions is equal to the integral of the right function minus the left function between the y-values of their intersection points, so it is necessary to determine which are the right and left functions and to find the intersection points.
The quadratic function has a positive leading coefficient, so the parabola opens to the right. All the choices are finite values, so the linear function must cross to the right of the quadratic (graph). Set the expressions equal and solve for y to find where they intersect.
| y2 − y + 1 = y + 1 | Set functions equal |
| y2 − 2y = 0 | Subtract y and 1 from both sides |
| y(y − 2) = 0 | Factor out GCF |
| y = 0 and y = 2 | Apply zero product property |
Therefore, the right function is f(y) = y + 1 and the left function is g(y) = y2 − y + 1, and they intersect at y = 0 and y = 2.
Integrate the difference between the functions on the interval of their intersection points to find the area of region R.
| Area of region R | |
| Substitute functions and values | |
| Combine like terms | |
| Integrate | |
| Apply FTC | |
| Simplify terms | |
| Subtract |
Therefore, the area of region R is .
Question
The graphs of the functions f (x) = 2x3 − x + 2 and g (x) = −2x3 + 3x + 2 are shown above. The graphs of f and g intersect at x = −1, x = 0, and x = 1. What is the area between the graphs of f and g on [−1, 1] ?
| A. 0 | |
| B. 1 | |
| C. 2 | |
| D. 4 |
Explanation
To find the area between the graphs offandgon the interval [−1, 1], write two integrals that meet at the inner intersection point at x = 0.
The given graph shows that f (x) ≥ g (x) on [−1, 0], so the integrand of the integral on this interval is f (x) − g (x). Write the integral for the first interval and evaluate.
| Area between f and g on [a, b] | |
| Substitute functions, a = -1, and b = 0 | |
| Combine like terms | |
| Integrate | |
| Apply FTC | |
| Simplify | |
| Subtract |
The area on the first interval [−1, 0] is 1.
The given graph shows that g (x) ≥ f (x) on [0, 1], so the integrand of the integral on this interval is g (x) −f (x). Write the integral for the second interval and evaluate.
| Area between f and g on [a, b] | |
| Substitute functions, a = 0, and b = 1 | |
| Combine like terms | |
| Integrate | |
| Apply FTC | |
| Simplify | |
| Subtract |
The area on the second interval [0, 1] is also 1.
The total area between the graphs of the given functions on [−1, 1] is the sum of the values of the two integrals, so the total area is 1 + 1 = 2.
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