Area Between Curves: Functions of x
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AP Calculus AB › Area Between Curves: Functions of x
For $2\le x\le 5$, $f(x)=\ln x+3$ and $g(x)=2$. Which integral gives the area between the curves?
$\displaystyle \int_{2}^{5}\big[2-(\ln x+3)\big]dx$
$\displaystyle \int_{2}^{5}\big[(\ln x+3)-2\big]dx$
$\displaystyle \int_{5}^{2}\big[(\ln x+3)-2\big]dx$
$\displaystyle \int_{0}^{5}\big[(\ln x+3)-2\big]dx$
$\displaystyle \int_{2}^{5}\big[(\ln x+3)+2\big]dx$
Explanation
This problem asks for the area between two curves, requiring integration of the difference between upper and lower functions. Since ln(x) is positive for x>1, we have f(x)=ln(x)+3 > 3 on [2,5], while g(x)=2 is constant, so f(x) is above g(x) throughout the interval. The area formula is ∫[upper-lower]dx, giving us ∫₂⁵[(ln x+3)-2]dx. Choice B reverses the subtraction to 2-(ln x+3), which would yield a negative result since ln x+3 > 2 on this interval. Choice D incorrectly starts at x=0, but ln(x) is undefined at x=0, making this integral improper. When finding area between curves, ensure the integrand represents (upper function minus lower function) over the valid domain.
For $-3\le x\le -1$, $f(x)=\frac{1}{x}+5$ lies above $g(x)=4$; which integral gives the area between them?
$\displaystyle \int_{-3}^{-1}\big((\tfrac{1}{x}+5)+4\big),dx$
$\displaystyle \int_{-3}^{-1}\big((\tfrac{1}{x}+5)-4\big),dx$
$\displaystyle \int_{-1}^{-3}\big(4-(\tfrac{1}{x}+5)\big),dx$
$\displaystyle \int_{-1}^{-3}\big((\tfrac{1}{x}+5)-4\big),dx$
$\displaystyle \int_{-3}^{-1}\big(4-(\tfrac{1}{x}+5)\big),dx$
Explanation
This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = 1/x + 5 is the upper curve and g(x) = 4 is the lower curve on [-3, -1], so the integrand is (1/x + 5) - 4. Integrating from -3 to -1 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice A, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
Between $x=-1$ and $x=2$, $f(x)=x^2+2x+3$ is above $g(x)=x+1$; which integral represents the area?
$\displaystyle \int_{-1}^{2}\big((x+1)-(x^2+2x+3)\big),dx$
$\displaystyle \int_{-1}^{2}\big((x^2+2x+3)+(x+1)\big),dx$
$\displaystyle \int_{2}^{-1}\big((x^2+2x+3)-(x+1)\big),dx$
$\displaystyle \int_{-1}^{2}\big((x^2+2x+3)-(x+1)\big),dx$
$\displaystyle \int_{2}^{-1}\big((x+1)-(x^2+2x+3)\big),dx$
Explanation
This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = x² + 2x + 3 is the upper curve and g(x) = x + 1 is the lower curve on [-1, 2], so the integrand is (x² + 2x + 3) - (x + 1). Integrating from -1 to 2 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice C, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
Between $x=0$ and $x=2$, $f(x)=3^x$ is above $g(x)=2^x$; which integral represents the area between them?
$\displaystyle \int_{2}^{0}\big(3^x-2^x\big),dx$
$\displaystyle \int_{0}^{2}\big(2^x-3^x\big),dx$
$\displaystyle \int_{0}^{2}\big(3^x-2^x\big),dx$
$\displaystyle \int_{0}^{2}\big(3^x+2^x\big),dx$
$\displaystyle \int_{2}^{0}\big(2^x-3^x\big),dx$
Explanation
This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = $3^x$ is the upper curve and g(x) = $2^x$ is the lower curve on [0, 2], so the integrand is $3^x$ - $2^x$. Integrating from 0 to 2 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice A, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
Between $x=-2$ and $x=2$, $f(x)=x^2+2$ lies above $g(x)=|x|$; which integral gives the area setup?
$\displaystyle \int_{-2}^{2}\big((x^2+2)+|x|\big),dx$
$\displaystyle \int_{2}^{-2}\big((x^2+2)-|x|\big),dx$
$\displaystyle \int_{-2}^{2}\big(|x|-(x^2+2)\big),dx$
$\displaystyle \int_{-2}^{2}\big((x^2+2)-|x|\big),dx$
$\displaystyle \int_{2}^{-2}\big(|x|-(x^2+2)\big),dx$
Explanation
This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = x² + 2 is the upper curve and g(x) = |x| is the lower curve on [-2, 2], so the integrand is (x² + 2) - |x|. Integrating from -2 to 2 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice A, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
For $1\le x\le 4$, $f(x)=\frac{x^2+1}{2}$ is above $g(x)=x$; which integral represents the area?
$\displaystyle \int_{4}^{1}\big(x-\tfrac{x^2+1}{2}\big),dx$
$\displaystyle \int_{1}^{4}\big(x-\tfrac{x^2+1}{2}\big),dx$
$\displaystyle \int_{1}^{4}\big(\tfrac{x^2+1}{2}-x\big),dx$
$\displaystyle \int_{1}^{4}\big(\tfrac{x^2+1}{2}+x\big),dx$
$\displaystyle \int_{4}^{1}\big(\tfrac{x^2+1}{2}-x\big),dx$
Explanation
This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = (x² + 1)/2 is the upper curve and g(x) = x is the lower curve on [1, 4], so the integrand is (x² + 1)/2 - x. Integrating from 1 to 4 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice B, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
On $1,5$, $f(x)=\frac{5}{x}+1$ lies above $g(x)=1$; which integral gives the area between the curves?
$\displaystyle \int_{5}^{1}\big(1-(\tfrac{5}{x}+1)\big),dx$
$\displaystyle \int_{1}^{5}\big((\tfrac{5}{x}+1)+1\big),dx$
$\displaystyle \int_{5}^{1}\big((\tfrac{5}{x}+1)-1\big),dx$
$\displaystyle \int_{1}^{5}\big((\tfrac{5}{x}+1)-1\big),dx$
$\displaystyle \int_{1}^{5}\big(1-(\tfrac{5}{x}+1)\big),dx$
Explanation
This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = 5/x + 1 is the upper curve and g(x) = 1 is the lower curve on [1, 5], so the integrand is (5/x + 1) - 1. Integrating from 1 to 5 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice B, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
Between $x=2$ and $x=6$, $f(x)=\sqrt{x}$ lies above $g(x)=\frac{x}{4}$; which integral gives the area between them?
$\displaystyle \int_{6}^{2}\big(\sqrt{x}-\tfrac{x}{4}\big),dx$
$\displaystyle \int_{2}^{6}\big(\sqrt{x}+\tfrac{x}{4}\big),dx$
$\displaystyle \int_{6}^{2}\big(\tfrac{x}{4}-\sqrt{x}\big),dx$
$\displaystyle \int_{2}^{6}\big(\sqrt{x}-\tfrac{x}{4}\big),dx$
$\displaystyle \int_{2}^{6}\big(\tfrac{x}{4}-\sqrt{x}\big),dx$
Explanation
This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = √x is the upper curve and g(x) = x/4 is the lower curve on [2, 6], so the integrand is √x - x/4. Integrating from 2 to 6 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice A, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
Between $x=-1$ and $x=3$, $f(x)=x+5$ is above $g(x)=2x+1$; which integral represents the area between them?
$\displaystyle \int_{-1}^{3}\big((x+5)-(2x+1)\big),dx$
$\displaystyle \int_{-1}^{3}\big((2x+1)-(x+5)\big),dx$
$\displaystyle \int_{3}^{-1}\big((x+5)-(2x+1)\big),dx$
$\displaystyle \int_{3}^{-1}\big((2x+1)-(x+5)\big),dx$
$\displaystyle \int_{-1}^{3}\big((x+5)+(2x+1)\big),dx$
Explanation
This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = x + 5 is the upper curve and g(x) = 2x + 1 is the lower curve on [-1, 3], so the integrand is (x + 5) - (2x + 1). Integrating from -1 to 3 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice B, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
Between $x=-\pi$ and $x=0$, $f(x)=3-\cos x$ is above $g(x)=1+\sin x$; which integral gives the area?
$\displaystyle \int_{0}^{-\pi}\big((1+\sin x)-(3-\cos x)\big),dx$
$\displaystyle \int_{0}^{-\pi}\big((3-\cos x)-(1+\sin x)\big),dx$
$\displaystyle \int_{-\pi}^{0}\big((3-\cos x)-(1+\sin x)\big),dx$
$\displaystyle \int_{-\pi}^{0}\big((1+\sin x)-(3-\cos x)\big),dx$
$\displaystyle \int_{-\pi}^{0}\big((3-\cos x)+(1+\sin x)\big),dx$
Explanation
This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = 3 - cos x is the upper curve and g(x) = 1 + sin x is the lower curve on [-π, 0], so the integrand is (3 - cos x) - (1 + sin x). Integrating from -π to 0 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice C, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.