Area Between Curves (x-functions) - Calculus 1

A region is bounded by the graphs of $y = 4 - x^2$ and $y = x$ . Which of the following integrals represents the area of this region?
Answer: $\int_{\frac{-1 - \sqrt{17}}{2}}^{\frac{-1 + \sqrt{17}}{2}} (4 - x - x^2) dx$ First, find the points of intersection by setting the two functions equal: $4 - x^2 = x$ . This rearranges to the quadratic equation $x^2 + x - 4 = 0$ . Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , the intersection points are $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-4)}}{2} = \frac{-1 \pm \sqrt{17}}{2}$ . These are the limits of integration. Next, determine which function is on top in the interval $(\frac{-1 - \sqrt{17}}{2}, \frac{-1 + \sqrt{17}}{2})$ . We can test a point in this interval, such as $x=0$ . At $x=0$ , $y=4-x^2=4$ and $y=x=0$ . Since $4 > 0$ , the parabola $y = 4-x^2$ is the upper function. The area is given by the integral of the upper function minus the lower function over the interval defined by the intersection points. Thus, the area is $\int_{\frac{-1 - \sqrt{17}}{2}}^{\frac{-1 + \sqrt{17}}{2}} ((4 - x^2) - x) dx = \int_{\frac{-1 - \sqrt{17}}{2}}^{\frac{-1 + \sqrt{17}}{2}} (4 - x - x^2) dx$ .
What is the area of the region bounded by $y = \sec^2(x)$ , $y = \sin(x)$ , $x = -\pi/4$ , and $x = \pi/4$ ?
Answer: 2 The area is given by the integral $A = \int_{-\pi/4}^{\pi/4} (\sec^2(x) - \sin(x)) dx$ , as $\sec^2(x) \ge 1$ and $\sin(x) \le 1$ on this interval, ensuring $\sec^2(x)$ is the upper function. We can split the integral: $A = \int_{-\pi/4}^{\pi/4} \sec^2(x) dx - \int_{-\pi/4}^{\pi/4} \sin(x) dx$ . The function $\sec^2(x)$ is an even function, so $\int_{-a}^{a} f(x)dx = 2\int_{0}^{a} f(x)dx$ . The function $\sin(x)$ is an odd function, so $\int_{-a}^{a} f(x)dx = 0$ . Therefore, $A = 2\int_{0}^{\pi/4} \sec^2(x) dx - 0$ . The antiderivative of $\sec^2(x)$ is $\tan(x)$ . So, $A = 2[\tan(x)]_{0}^{\pi/4} = 2(\tan(\pi/4) - \tan(0)) = 2(1 - 0) = 2$ .
Let R be the region in the first quadrant bounded by the graphs of $y = e^x$ , $y = e^{-x}$ , and the line $x = \ln(2)$ . What is the area of R?
Answer: $1/2$ The region R is bounded by $x=0$ (first quadrant), $x=\ln(2)$ , $y=e^x$ , and $y=e^{-x}$ . In the first quadrant, for $x>0$ , $e^x > e^{-x}$ . The area is given by the integral $A = \int_{0}^{\ln(2)} (e^x - e^{-x}) dx$ . The antiderivative is $e^x - (-e^{-x}) = e^x + e^{-x}$ . Evaluating using the Fundamental Theorem of Calculus: $[e^x + e^{-x}]_{0}^{\ln(2)} = (e^{\ln(2)} + e^{-\ln(2)}) - (e^0 + e^0) = (2 + \frac{1}{2}) - (1 + 1) = \frac{5}{2} - 2 = \frac{1}{2}$ .
Find the area of the region bounded by the curves $y = |x|$ and $y = 2 - x^2$ .
Answer: $7/3$ The region is symmetric with respect to the y-axis because both $y=|x|$ and $y=2-x^2$ are even functions. We can find the area for $x \ge 0$ and double it. First, find the intersection points. For $x \ge 0$ , we solve $x = 2 - x^2$ , which gives $x^2 + x - 2 = 0$ , or $(x+2)(x-1) = 0$ . Since $x \ge 0$ , the intersection is at $x=1$ . By symmetry, the other intersection is at $x=-1$ . For $x \in [0, 1]$ , $2-x^2 \ge x$ . The area in the first quadrant is $A_1 = \int_{0}^{1} ((2 - x^2) - x) dx$ . Evaluating the integral: $A_1 = [2x - \frac{x^3}{3} - \frac{x^2}{2}]_{0}^{1} = (2 - \frac{1}{3} - \frac{1}{2}) - 0 = \frac{12-2-3}{6} = \frac{7}{6}$ . The total area is twice this value due to symmetry: $A = 2 \cdot A_1 = 2 \cdot \frac{7}{6} = \frac{7}{3}$ .
What is the total area of the finite region(s) bounded by the curve $y = x\sqrt{1-x^2}$ and the x-axis?
Answer: $2/3$ The domain of $y = x\sqrt{1-x^2}$ is $[-1, 1]$ . The curve intersects the x-axis at $x=-1, 0, 1$ . The function is positive for $x \in (0,1]$ and negative for $x \in [-1,0)$ . Since this is an odd function ( $f(-x) = -f(x)$ ), the areas of the two regions are equal. We can calculate the area of the region for $x \in [0,1]$ and double it. Area of one region: $A_1 = \int_{0}^{1} x\sqrt{1-x^2} dx$ . Use u-substitution: let $u = 1-x^2$ , so $du = -2x dx$ , or $-du/2 = x dx$ . When $x=0, u=1$ . When $x=1, u=0$ . The integral becomes $\int_{1}^{0} \sqrt{u} (-\frac{1}{2} du) = \frac{1}{2} \int_{0}^{1} u^{1/2} du = \frac{1}{2} [\frac{2}{3}u^{3/2}]_{0}^{1} = \frac{1}{2} (\frac{2}{3} - 0) = \frac{1}{3}$ . The total area is the sum of the areas of the two regions, so $A_{total} = A_1 + A_2 = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$ .
The area of the region bounded by $y = e^x$ , $y=1$ and $x=2$ is:
Answer: $e^2 - 3$ The curve $y=e^x$ intersects $y=1$ at $x=0$ . The region is bounded by $x=0$ and $x=2$ . In this interval, $e^x \ge 1$ . The area is given by $A = \int_{0}^{2} (e^x - 1) dx$ . The antiderivative is $e^x - x$ . Evaluating the definite integral: $A = [e^x - x]_{0}^{2} = (e^2 - 2) - (e^0 - 0) = (e^2 - 2) - (1) = e^2 - 3$ .
Which of the following expressions represents the area of the region enclosed by the graphs of $y = \sin(x)$ and $y = \cos(x)$ for $x \in [0, \pi/2]$ ?
Answer: $\int_{0}^{\pi/4} (\cos(x) - \sin(x)) dx + \int_{\pi/4}^{\pi/2} (\sin(x) - \cos(x)) dx$ To find the area between two curves, we must integrate the upper function minus the lower function. The functions $\sin(x)$ and $\cos(x)$ intersect when $\sin(x) = \cos(x)$ , which occurs at $x = \pi/4$ in the interval $[0, \pi/2]$ . On the interval $[0, \pi/4]$ , $\cos(x) \ge \sin(x)$ . On the interval $[\pi/4, \pi/2]$ , $\sin(x) \ge \cos(x)$ . Therefore, the total area requires splitting the integral at the intersection point: $Area = \int_{0}^{\pi/4} (\text{upper} - \text{lower}) dx + \int_{\pi/4}^{\pi/2} (\text{upper} - \text{lower}) dx = \int_{0}^{\pi/4} (\cos(x) - \sin(x)) dx + \int_{\pi/4}^{\pi/2} (\sin(x) - \cos(x)) dx$ .
Let R be the region bounded by $y = \cos(x)$ , the x-axis, $x = -\pi/2$ , and $x = \pi/2$ . A vertical line $x=c$ is drawn such that it divides R into two regions of equal area. What must be true about $c$ ?
Answer: $c=0$ The function $y = \cos(x)$ is an even function, meaning its graph is symmetric with respect to the y-axis. The interval $[-\pi/2, \pi/2]$ is also symmetric about $x=0$ . Because of this symmetry, the total area of the region R is $A = \int_{-\pi/2}^{\pi/2} \cos(x) dx = 2 \int_{0}^{\pi/2} \cos(x) dx$ . The line that divides this symmetric region into two equal areas must be the line of symmetry itself, which is the y-axis, or $x=0$ . Therefore, $c=0$ .
What is the area of the finite region bounded by the curve $y = x^2$ , its tangent line at $x = 1$ , and the y-axis?
Answer: $1/3$ First, find the equation of the tangent line. Let $f(x) = x^2$ . The derivative is $f'(x) = 2x$ . At $x = 1$ , the y-coordinate is $f(1) = 1^2 = 1$ and the slope of the tangent line is $m = f'(1) = 2(1) = 2$ . The equation of the tangent line is $y - 1 = 2(x - 1)$ , which simplifies to $y = 2x - 1$ . The region is bounded by $y = x^2$ , $y = 2x - 1$ , and the y-axis ( $x=0$ ). The intersection is at $x=1$ . We need to find the area between these curves from $x = 0$ to $x = 1$ . For $x \in [0, 1]$ , $x^2 \ge 2x - 1$ because $x^2 - (2x-1) = (x-1)^2 \ge 0$ . The area is $A = \int_{0}^{1} (x^2 - (2x - 1)) dx = \int_{0}^{1} (x^2 - 2x + 1) dx$ . Evaluating the integral: $[\frac{x^3}{3} - x^2 + x]_{0}^{1} = (\frac{1}{3} - 1 + 1) - 0 = \frac{1}{3}$ .
Let $A_1$ be the area of the region bounded by $y=x^2$ and $y=x$ . Let $A_2$ be the area of the region bounded by $y=\sqrt{x}$ and $y=x$ . What is the value of the ratio $A_1/A_2$ ?
Answer: $1$ First, calculate $A_1$ . The intersection points of $y=x^2$ and $y=x$ are found by solving $x^2=x$ , which gives $x(x-1)=0$ , so $x=0$ and $x=1$ . On the interval $[0,1]$ , $x \ge x^2$ . $A_1 = \int_0^1 (x - x^2) dx = [\frac{x^2}{2} - \frac{x^3}{3}]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ . Next, calculate $A_2$ . The intersection points of $y=\sqrt{x}$ and $y=x$ are found by solving $\sqrt{x}=x$ , which gives $x=x^2$ , so $x(x-1)=0$ , and $x=0, 1$ . On the interval $[0,1]$ , $\sqrt{x} \ge x$ . $A_2 = \int_0^1 (\sqrt{x} - x) dx = [\frac{2}{3}x^{3/2} - \frac{x^2}{2}]_0^1 = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6}$ . The ratio is $A_1/A_2 = (1/6)/(1/6) = 1$ . Conceptually, the functions $y=x^2$ (for $x\ge 0$ ) and $y=\sqrt{x}$ are inverses of each other. The area between $y=x$ and $y=x^2$ is a reflection across the line $y=x$ of the area between $y=x$ and $y=\sqrt{x}$ . Since reflection is an isometry, the areas must be equal.
Let $f(x) = x^3$ . Let $L$ be the line that is tangent to the graph of $f$ at $x = 1$ . What is the area of the region enclosed by the graph of $f$ , the line $L$ , and the line $x = -1$ ?
Answer: 4 First, find the equation of the tangent line $L$ . $f(x) = x^3 \Rightarrow f'(x) = 3x^2$ . At $x = 1$ , $f(1) = 1$ and the slope is $m = f'(1) = 3$ . The tangent line is $y - 1 = 3(x - 1) \Rightarrow y = 3x - 2$ . The region is bounded by $y = x^3$ , $y = 3x - 2$ , and $x = -1$ . The tangent point is at $x=1$ . The area is $A = \int_{-1}^{1} |x^3 - (3x - 2)| dx$ . On the interval $[-1, 1]$ , the cubic function $y=x^3$ is above the tangent line $y=3x-2$ . For instance, at $x=0$ , $0 > -2$ . At $x=-1$ , $-1 > -5$ . So the integrand is $x^3 - 3x + 2$ . $A = \int_{-1}^{1} (x^3 - 3x + 2) dx$ . The antiderivative is $[\frac{x^4}{4} - \frac{3x^2}{2} + 2x]_{-1}^{1}$ . Evaluating: $(\frac{1}{4} - \frac{3}{2} + 2) - (\frac{1}{4} - \frac{3}{2} - 2) = (\frac{1-6+8}{4}) - (\frac{1-6-8}{4}) = \frac{3}{4} - (-\frac{13}{4}) = \frac{16}{4} = 4$ .
A region is bounded by $y = \sqrt{x}$ , the x-axis, and the line $x = 4$ . A vertical line $x=k$ divides this region into two smaller regions of equal area. What is the value of $k$ ?
Answer: $4^{2/3}$ First, find the total area of the region. $A_{total} = \int_{0}^{4} \sqrt{x} dx = \int_{0}^{4} x^{1/2} dx = [\frac{2}{3}x^{3/2}]_{0}^{4} = \frac{2}{3}(4^{3/2}) - 0 = \frac{2}{3}(8) = \frac{16}{3}$ . We want to find $k$ such that the area from 0 to $k$ is half the total area, which is $\frac{1}{2} \cdot \frac{16}{3} = \frac{8}{3}$ . Set up the integral for the area of the left subregion: $A_{left} = \int_{0}^{k} \sqrt{x} dx = [\frac{2}{3}x^{3/2}]_{0}^{k} = \frac{2}{3}k^{3/2}$ . Now, set this equal to half the total area: $\frac{2}{3}k^{3/2} = \frac{8}{3} \Rightarrow k^{3/2} = 4 \Rightarrow k = 4^{2/3}$ .
What is the area of the region bounded by $y = \frac{1}{x}$ , $y = x$ , and $x=2$ ?
Answer: $\frac{3}{2} - \ln(2)$ The curves $y = 1/x$ and $y = x$ intersect when $x = 1/x \Rightarrow x^2 = 1$ . In the first quadrant, this is $x=1$ . The region is bounded by $x=1$ and $x=2$ . In the interval $[1, 2]$ , $x \ge 1/x$ . The area is given by $A = \int_{1}^{2} (x - \frac{1}{x}) dx$ . The antiderivative is $\frac{x^2}{2} - \ln|x|$ . Evaluating from 1 to 2: $[\frac{x^2}{2} - \ln(x)]_{1}^{2} = (\frac{2^2}{2} - \ln(2)) - (\frac{1^2}{2} - \ln(1)) = (2 - \ln(2)) - (\frac{1}{2} - 0) = 2 - \frac{1}{2} - \ln(2) = \frac{3}{2} - \ln(2)$ .
A region is bounded by the curves $y = \frac{2}{1+x^2}$ and $y=1$ . Which integral represents the area of this region?
Answer: $\int_{-1}^{1} (\frac{2}{1+x^2} - 1) dx$ First, find the points of intersection by setting the functions equal: $1 = \frac{2}{1+x^2}$ . This gives $1+x^2 = 2$ , so $x^2=1$ , and the intersections are at $x=-1$ and $x=1$ . These will be the limits of integration. Next, determine which function is the upper boundary on the interval $[-1,1]$ . Choose a test point within the interval, like $x=0$ . At $x=0$ , $y=\frac{2}{1+0^2} = 2$ , while the other function is $y=1$ . Since $2 > 1$ , the function $y = \frac{2}{1+x^2}$ is the upper function. The area is found by integrating the difference (upper function - lower function) between the points of intersection. Area $= \int_{-1}^{1} (\frac{2}{1+x^2} - 1) dx$ .
Let R be the region bounded by $y=x^2-2x$ and $y=x$ . Which of the following integrals represents the area of R?
Answer: $\int_{0}^{3} (3x-x^2) dx$ When finding the area between two curves, you need to determine which function is on top, find their intersection points, and set up the integral as $\int_a^b [\text{top function} - \text{bottom function}] \, dx$ . First, find where $y = x^2 - 2x$ and $y = x$ intersect by setting them equal: $x^2 - 2x = x$ , which gives $x^2 - 3x = 0$ , so $x(x-3) = 0$ . The curves intersect at $x = 0$ and $x = 3$ . Next, determine which function is on top in the interval $[0,3]$ . Test a point like $x = 1$ : the line gives $y = 1$ , while the parabola gives $y = 1 - 2 = -1$ . Since $1 > -1$ , the line $y = x$ is above the parabola $y = x^2 - 2x$ throughout this interval. The area is therefore $\int_0^3 [x - (x^2 - 2x)] \, dx = \int_0^3 (3x - x^2) \, dx$ , which matches answer choice D. Looking at the wrong answers: A uses the wrong integrand $x^2 - 3x$ (parabola minus line instead of line minus parabola). B incorrectly splits the integral and adds areas of individual functions rather than their difference. C has the correct integrand but wrong limits—it uses $-1$ to 3 instead of 0 to 3. Study tip: Always sketch the curves or test a point between intersections to determine which function is on top. The area between curves is always "top minus bottom," and getting this backwards is the most common error.
The graphs of $y = 2x$ and $y = x^2$ intersect at two points. What is the area of the region enclosed between them?
Answer: $4/3$ First, find the points of intersection by setting the equations equal: $x^2 = 2x \Rightarrow x^2 - 2x = 0 \Rightarrow x(x-2) = 0$ . The intersection points are at $x=0$ and $x=2$ . In the interval $[0, 2]$ , the line $y=2x$ is above the parabola $y=x^2$ (e.g., at $x=1$ , $2 > 1$ ). The area is given by the integral $A = \int_{0}^{2} (2x - x^2) dx$ . The antiderivative is $x^2 - \frac{x^3}{3}$ . Evaluating the integral: $A = [x^2 - \frac{x^3}{3}]_{0}^{2} = (2^2 - \frac{2^3}{3}) - 0 = 4 - \frac{8}{3} = \frac{12-8}{3} = \frac{4}{3}$ .
The line $x=c$ divides the area of the region bounded by $y=2x-x^2$ and the x-axis into two equal halves. Which equation must be satisfied by $c$ ?
Answer: $\int_{0}^{c} (2x-x^2) dx = \frac{2}{3}$ First, find the total area of the region. The curve $y=2x-x^2 = x(2-x)$ intersects the x-axis at $x=0$ and $x=2$ . The total area is $A = \int_{0}^{2} (2x - x^2) dx = [x^2 - \frac{x^3}{3}]_{0}^{2} = (4 - \frac{8}{3}) - 0 = \frac{4}{3}$ . We want the line $x=c$ to divide this area in half, so the area of the region from $x=0$ to $x=c$ must be $A/2 = (4/3)/2 = 2/3$ . This is represented by the equation $\int_{0}^{c} (2x-x^2) dx = \frac{2}{3}$ .
What is the area of the region bounded by the graphs of $f(x)=|x|$ and $g(x)=-x^2+2$ ?
Answer: $7/3$ When finding the area between two curves, you need to identify where they intersect, determine which function is on top, and set up the appropriate integral. First, find the intersection points by solving $|x| = -x^2 + 2$ . Since we're dealing with absolute value, consider two cases. For $x \geq 0$ : $x = -x^2 + 2$ , so $x^2 + x - 2 = 0$ , giving $(x+2)(x-1) = 0$ . Since $x \geq 0$ , we get $x = 1$ . For $x < 0$ : $-x = -x^2 + 2$ , so $x^2 - x - 2 = 0$ , giving $(x-2)(x+1) = 0$ . Since $x < 0$ , we get $x = -1$ . The curves intersect at $(-1, 1)$ and $(1, 1)$ . Between $x = -1$ and $x = 1$ , the parabola $g(x) = -x^2 + 2$ is above $f(x) = |x|$ . The area is: $\int_{-1}^{1} [(-x^2 + 2) - |x|] \, dx$ Since $|x| = -x$ for $x \in [-1, 0]$ and $|x| = x$ for $x \in [0, 1]$ : $\int_{-1}^{0} [(-x^2 + 2) - (-x)] \, dx + \int_{0}^{1} [(-x^2 + 2) - x] \, dx$ $= \int_{-1}^{0} (-x^2 + x + 2) \, dx + \int_{0}^{1} (-x^2 - x + 2) \, dx$ $= \left[-\frac{x^3}{3} + \frac{x^2}{2} + 2x\right]_{-1}^{0} + \left[-\frac{x^3}{3} - \frac{x^2}{2} + 2x\right]_{0}^{1}$ $= \frac{7}{6} + \frac{7}{6} = \frac{7}{3}$ Answer (A) 1 likely comes from forgetting the absolute value. Answer (B) $\frac{7}{6}$ is just half the correct answer. Answer (C) $\frac{10}{3}$ might result from calculation errors with the absolute value. Always sketch both functions first to visualize which is on top in each interval—this prevents sign errors in your integral setup.
Find the area of the region bounded by $y = 12 - x^2$ and $y = x^2 - 6$ .
Answer: 72 First, find the points of intersection: $12 - x^2 = x^2 - 6 \Rightarrow 18 = 2x^2 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$ . The upper curve is $y = 12 - x^2$ and the lower curve is $y = x^2 - 6$ . The area is $A = \int_{-3}^{3} ((12 - x^2) - (x^2 - 6)) dx = \int_{-3}^{3} (18 - 2x^2) dx$ . By symmetry, $A = 2 \int_{0}^{3} (18 - 2x^2) dx = 2[18x - \frac{2x^3}{3}]_{0}^{3} = 2((18(3) - \frac{2(3^3)}{3}) - 0) = 2(54 - \frac{54}{3}) = 2(54 - 18) = 2(36) = 72$ .
What is the area of the region in the first quadrant bounded on the left by the y-axis, above by $y = \cos(x)$ , and below by $y = \sin(x)$ ?
Answer: $\sqrt{2} - 1$ The region is bounded on the left by the y-axis ( $x=0$ ). The upper and lower boundaries $y=\cos(x)$ and $y=\sin(x)$ intersect when $\cos(x)=\sin(x)$ , which in the first quadrant is at $x=\pi/4$ . So the region extends from $x=0$ to $x=\pi/4$ . In this interval, $\cos(x) \ge \sin(x)$ . The area is $A = \int_{0}^{\pi/4} (\cos(x) - \sin(x)) dx$ . The antiderivative is $\sin(x) - (-\cos(x)) = \sin(x) + \cos(x)$ . Evaluating the integral: $A = [\sin(x) + \cos(x)]_{0}^{\pi/4} = (\sin(\pi/4) + \cos(\pi/4)) - (\sin(0) + \cos(0)) = (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) - (0 + 1) = \sqrt{2} - 1$ .

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