AP Calculus AB: Area Between Curves (Drill 1)

Question 1. The area of the region enclosed by \( y = 2x + 3 \) and \( y = x^2 \) is:

• A) \( \dfrac{32}{3} \) ✓
• B) \( 20 \)
• C) \( 9 \)
• D) \( \dfrac{16}{3} \)

Explanation: Choice A is correct. Set \( 2x+3 = x^2 \): \( x^2-2x-3 = 0 \), so \( (x-3)(x+1) = 0 \), giving \( x = -1 \) and \( x = 3 \). On \( [-1, 3] \) the line lies above the parabola. Area \( = \int_{-1}^{3}[(2x+3)-x^2]\,dx = \left[-\dfrac{x^3}{3}+x^2+3x\right]_{-1}^{3} = 9 - \left(-\dfrac{7}{3}\right) = \dfrac{32}{3} \). Choice B is incorrect because the student integrated only the top curve \( \int_{-1}^3(2x+3)\,dx = 20 \), forgetting to subtract the lower curve \( x^2 \). Choice C is incorrect because the student used bounds \( [0, 3] \) instead of \( [-1, 3] \), perhaps discarding the negative root when solving for intersection points, and missed the portion of the region to the left of the y-axis: \( \int_0^3[(2x+3)-x^2]\,dx = 9 \). Choice D is incorrect because the student correctly set up the integrand but used the vertex of the parabola at \( x = 1 \) as the lower bound of integration rather than the intersection point at \( x = -1 \), computing \( \int_1^3[(-x^2+2x+3)]\,dx = \dfrac{16}{3} \) and missing the left portion of the enclosed region.

Question 2. The area of the region enclosed by \( y = \sqrt{x} \) and \( y = \dfrac{x}{2} \) is:

• A) \( \dfrac{4}{3} \) ✓
• B) \( \dfrac{16}{3} \)
• C) \( \dfrac{8}{3} \)
• D) \( 4 \)

Explanation: Choice A is correct. Set \( \sqrt{x} = \dfrac{x}{2} \): for \( x \ge 0 \), squaring gives \( x = \dfrac{x^2}{4} \), so \( x^2 - 4x = 0 \) and \( x = 0 \) or \( x = 4 \). On \( [0,4] \), \( \sqrt{x} \ge \dfrac{x}{2} \). Area \( = \int_0^4\!\left(\sqrt{x} - \dfrac{x}{2}\right)dx = \left[\dfrac{2x^{3/2}}{3} - \dfrac{x^2}{4}\right]_0^4 = \dfrac{16}{3} - 4 = \dfrac{4}{3} \). Choice B is incorrect because the student integrated only the upper curve \( \sqrt{x} \) and forgot to subtract \( \dfrac{x}{2} \): \( \int_0^4 \sqrt{x}\,dx = \dfrac{16}{3} \). Choice C is incorrect because the student found the antiderivative of \( \dfrac{x}{2} \) as \( \dfrac{x^2}{2} \) instead of \( \dfrac{x^2}{4} \), giving \( \left[\dfrac{2x^{3/2}}{3}-\dfrac{x^2}{2}\right]_0^4 = \dfrac{16}{3}-8 = -\dfrac{8}{3} \) and taking the absolute value to arrive at \( \dfrac{8}{3} \). Choice D is incorrect because the student found the antiderivative of \( \sqrt{x} \) as \( \dfrac{x^2}{2} \), the antiderivative of \( x \), not \( x^{1/2} \), giving \( \int_0^4\!\left(x - \dfrac{x}{2}\right)dx = \int_0^4 \dfrac{x}{2}\,dx = 4 \).

Question 3. The region bounded by \( x = y^2 \) and \( x = 4 \) can be computed by integrating with respect to \( y \). Which of the following correctly expresses this area?

• A) \( \displaystyle\int_{-2}^{2}(4 - y^2)\,dy \) ✓
• B) \( \displaystyle\int_{0}^{2}(4 - y^2)\,dy \)
• C) \( \displaystyle\int_{-2}^{2}(y^2 - 4)\,dy \)
• D) \( \displaystyle\int_{0}^{4}(4 - y^2)\,dy \)

Explanation: Choice A is correct. The parabola \( x = y^2 \) and vertical line \( x = 4 \) intersect when \( y^2 = 4 \), giving \( y = \pm 2 \). When integrating with respect to \( y \), area is computed as \( \int(\text{right curve} - \text{left curve})\,dy \). Here the right boundary is \( x = 4 \) and the left boundary is \( x = y^2 \), so the integrand is \( 4 - y^2 \) and the limits run from \( y = -2 \) to \( y = 2 \). Choice B is incorrect because the student used only the upper half of the region, integrating from \( y = 0 \) to \( y = 2 \); this yields exactly half the total area and ignores the symmetric lower half. Choice C is incorrect because the student wrote the integrand as \( y^2 - 4 \) rather than \( 4 - y^2 \), subtracting the right boundary from the left instead of left from right, when integrating with respect to \( y \), the correct form is right minus left, and reversing the order produces a negative value. Choice D is incorrect because the student used the \( x \)-extent of the region, \( [0, 4] \), as the limits for the \( y \)-integral; the \( y \)-values run from \( -2 \) to \( 2 \), confusing the range of \( x \) with the range of \( y \).

Question 4. The area of the region enclosed by \( x = y^2 \) and \( x = y + 2 \) is:

• A) \( \dfrac{9}{2} \) ✓
• B) \( \dfrac{3}{2} \)
• C) \( \dfrac{10}{3} \)
• D) \( \dfrac{16}{3} \)

Explanation: Choice A is correct. Because both curves are most naturally expressed as functions of \( y \), integrating with respect to \( y \) requires only a single integral, whereas integrating with respect to \( x \) would require splitting the region, since the parabola has two branches. Find the intersections: \( y^2 = y+2 \Rightarrow y^2-y-2=0 \Rightarrow (y-2)(y+1)=0 \), so \( y=-1 \) and \( y=2 \). On \( [-1,2] \), the right curve is \( x = y+2 \) and the left is \( x = y^2 \). Area \( = \int_{-1}^{2}[(y+2)-y^2]\,dy = \left[\dfrac{y^2}{2}+2y-\dfrac{y^3}{3}\right]_{-1}^{2} = \dfrac{10}{3} - \left(-\dfrac{7}{6}\right) = \dfrac{20}{6}+\dfrac{7}{6} = \dfrac{9}{2} \). Choice B is incorrect because the student swapped the signs of the intersection solutions, using integration bounds \( [-2, 1] \) instead of \( [-1, 2] \), and obtained \( \dfrac{3}{2} \). Choice C is incorrect because the student used bounds \( [0, 2] \), starting at the vertex of the parabola rather than the lower intersection point, and computed only the upper portion of the enclosed area: \( \dfrac{10}{3} \). Choice D is incorrect because the student set up the integral with respect to \( x \), using only the upper branch \( y = \sqrt{x} \) and integrating \( \int_0^4\![\sqrt{x}-(x-2)]\,dx = \dfrac{16}{3} \), which accounts for only part of the region and misses the area below the \( x \)-axis between the lower branch and the line.

Question 5. The area of the region bounded by \( y = e^x \), \( y = 1 \), \( x = 0 \), and \( x = 2 \) is:

• A) \( e^2 - 3 \) ✓
• B) \( e^2 - 1 \)
• C) \( e^2 - 2 \)
• D) \( e^2 + 1 \)

Explanation: Choice A is correct. On \( [0,2] \), \( e^x \ge 1 \), so the area is \( \int_0^2(e^x-1)\,dx = \left[e^x-x\right]_0^2 = (e^2-2)-(e^0-0) = e^2-2-1 = e^2-3 \). Choice B is incorrect because the student integrated only \( e^x \) without subtracting the lower boundary \( y = 1 \): \( \int_0^2 e^x\,dx = e^2-1 \). Choice C is incorrect because the student evaluated the antiderivative \( e^x-x \) only at the upper limit \( x = 2 \), obtaining \( e^2-2 \), and forgot to subtract the value at the lower limit \( x = 0 \), which is \( e^0-0 = 1 \). Choice D is incorrect because the student wrote the antiderivative of \( -1 \) as \( +x \) instead of \( -x \), computing \( \left[e^x+x\right]_0^2 = (e^2+2)-1 = e^2+1 \).