Area Between Curves: Functions of x
Help Questions
AP Calculus AB › Area Between Curves: Functions of x
On $0,\pi$, $f(x)=1+\sin x$ is above $g(x)=\sin^2 x$; which integral represents the area between them?
$\displaystyle \int_{\pi}^{0}\big((1+\sin x)-\sin^2 x\big),dx$
$\displaystyle \int_{0}^{\pi}\big((1+\sin x)+\sin^2 x\big),dx$
$\displaystyle \int_{0}^{\pi}\big(\sin^2 x-(1+\sin x)\big),dx$
$\displaystyle \int_{\pi}^{0}\big(\sin^2 x-(1+\sin x)\big),dx$
$\displaystyle \int_{0}^{\pi}\big((1+\sin x)-\sin^2 x\big),dx$
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) = 1 + sin x is the upper curve and g(x) = sin² x is the lower curve on [0, π], so the integrand is (1 + sin x) - sin² x. Integrating from 0 to π 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 $x=0$ and $x=2$, $f(x)=3^x$ is above $g(x)=2^x$; which integral represents the area between them?
$\displaystyle \int_{2}^{0}\big(2^x-3^x\big),dx$
$\displaystyle \int_{2}^{0}\big(3^x-2^x\big),dx$
$\displaystyle \int_{0}^{2}\big(3^x+2^x\big),dx$
$\displaystyle \int_{0}^{2}\big(3^x-2^x\big),dx$
$\displaystyle \int_{0}^{2}\big(2^x-3^x\big),dx$
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.
For $1\le x\le 4$, $f(x)=\frac{x^2+1}{2}$ is above $g(x)=x$; which integral represents the area?
$\displaystyle \int_{4}^{1}\big(x-\tfrac{x^2+1}{2}\big),dx$
$\displaystyle \int_{1}^{4}\big(\tfrac{x^2+1}{2}-x\big),dx$
$\displaystyle \int_{4}^{1}\big(\tfrac{x^2+1}{2}-x\big),dx$
$\displaystyle \int_{1}^{4}\big(\tfrac{x^2+1}{2}+x\big),dx$
$\displaystyle \int_{1}^{4}\big(x-\tfrac{x^2+1}{2}\big),dx$
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 + 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.
Between $x=-1$ and $x=2$, $f(x)=x^2+2x+3$ is above $g(x)=x+1$; which integral represents the area?
$\displaystyle \int_{2}^{-1}\big((x^2+2x+3)-(x+1)\big),dx$
$\displaystyle \int_{-1}^{2}\big((x^2+2x+3)-(x+1)\big),dx$
$\displaystyle \int_{-1}^{2}\big((x+1)-(x^2+2x+3)\big),dx$
$\displaystyle \int_{-1}^{2}\big((x^2+2x+3)+(x+1)\big),dx$
$\displaystyle \int_{2}^{-1}\big((x+1)-(x^2+2x+3)\big),dx$
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.
For $0\le x\le 2$, $f(x)=x^2+1$ and $g(x)=3x+1$ with $g(x)\ge f(x)$. Which integral gives the area between them?
$\displaystyle \int_{0}^{2}\big[(3x+1)+(x^2+1)\big],dx$
$\displaystyle \int_{0}^{3}\big[(3x+1)-(x^2+1)\big],dx$
$\displaystyle \int_{0}^{2}\big[(3x+1)-(x^2+1)\big],dx$
$\displaystyle \int_{2}^{0}\big[(3x+1)-(x^2+1)\big],dx$
$\displaystyle \int_{0}^{2}\big[(x^2+1)-(3x+1)\big],dx$
Explanation
This problem requires finding the area between curves using the fundamental principle of integrating the upper function minus the lower function. Since we're told that g(x) ≥ f(x) on the interval [0, 2], this means g(x) = 3x + 1 is the upper function and f(x) = x² + 1 is the lower function. The area between curves is always calculated as ∫[upper - lower]dx, which gives us ∫₀²[(3x + 1) - (x² + 1)]dx. Choice A incorrectly subtracts in the wrong order, which would give a negative area. The key strategy is to always identify which function is on top before setting up the integral.
For $0\le x\le 1$, $f(x)=e^x$ is above $g(x)=1+x$. Which integral gives the area between them?
$\displaystyle \int_{0}^{1}\big[e^x-(1+x)\big]dx$
$\displaystyle \int_{0}^{1}\big[e^x+(1+x)\big]dx$
$\displaystyle \int_{0}^{e}\big[e^x-(1+x)\big]dx$
$\displaystyle \int_{0}^{1}\big[(1+x)-e^x\big]dx$
$\displaystyle \int_{1}^{0}\big[e^x-(1+x)\big]dx$
Explanation
To find the area between curves, we must integrate the upper function minus the lower function over the specified interval. Since f(x) = eˣ is above g(x) = 1 + x on [0, 1], the correct integral is ∫₀¹[eˣ - (1 + x)]dx. This follows the fundamental principle of subtracting the lower function from the upper function to obtain positive area. Choice A incorrectly computes [(1 + x) - eˣ], reversing the subtraction order and yielding a negative result. The strategy for any area problem is to first identify which function is on top, then set up the integral as upper minus lower.
For $-1\le x\le 1$, $f(x)=2-x^2$ is above $g(x)=x$. Which integral gives the area between the curves?
$\displaystyle \int_{1}^{-1}\big[(2-x^2)-x\big]dx$
$\displaystyle \int_{-1}^{1}\big[x-(2-x^2)\big]dx$
$\displaystyle \int_{-1}^{2}\big[(2-x^2)-x\big]dx$
$\displaystyle \int_{-1}^{1}\big[(2-x^2)-x\big]dx$
$\displaystyle \int_{-1}^{1}\big[(2-x^2)+x\big]dx$
Explanation
Finding the area between curves requires integrating the upper function minus the lower function over the specified interval. Given that f(x) = 2 - x² is above g(x) = x on [-1, 1], we need ∫₋₁¹[(2 - x²) - x]dx. This maintains the proper order of upper minus lower to ensure a positive area result. Choice A incorrectly places x first and then subtracts (2 - x²), which reverses the roles and would give the negative of the actual area. The fundamental principle is to always subtract in the order that preserves the geometric meaning of area as a positive quantity.
For $2\le x\le 5$, $f(x)=\ln x$ lies above $g(x)=\ln 2$. Which integral gives the area between the graphs?
$\displaystyle \int_{2}^{5}\big[\ln x+\ln 2\big]dx$
$\displaystyle \int_{2}^{5}\big[\ln x-\ln 2\big]dx$
$\displaystyle \int_{0}^{5}\big[\ln x-\ln 2\big]dx$
$\displaystyle \int_{5}^{2}\big[\ln x-\ln 2\big]dx$
$\displaystyle \int_{2}^{5}\big[(\ln 2)-\ln x\big]dx$
Explanation
To find the area between curves, we integrate the difference of the upper function minus the lower function. The problem states that f(x) = ln x lies above g(x) = ln 2 (a horizontal line) on [2, 5]. The correct setup is ∫₂⁵[ln x - ln 2]dx, subtracting the lower constant function from the upper logarithmic function. Choice A reverses this order to [ln 2 - ln x], which would give a negative result since ln x > ln 2 for x > 2. The key insight is that even when one function is constant, the same upper-minus-lower principle applies for finding enclosed area.
For $-3\le x\le -1$, $f(x)=\frac{1}{x}+5$ lies above $g(x)=4$; which integral gives the area between them?
$\displaystyle \int_{-3}^{-1}\big((\tfrac{1}{x}+5)+4\big),dx$
$\displaystyle \int_{-1}^{-3}\big((\tfrac{1}{x}+5)-4\big),dx$
$\displaystyle \int_{-1}^{-3}\big(4-(\tfrac{1}{x}+5)\big),dx$
$\displaystyle \int_{-3}^{-1}\big((\tfrac{1}{x}+5)-4\big),dx$
$\displaystyle \int_{-3}^{-1}\big(4-(\tfrac{1}{x}+5)\big),dx$
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.
On $1,5$, $f(x)=\frac{5}{x}+1$ lies above $g(x)=1$; which integral gives the area between the curves?
$\displaystyle \int_{5}^{1}\big((\tfrac{5}{x}+1)-1\big),dx$
$\displaystyle \int_{1}^{5}\big((\tfrac{5}{x}+1)+1\big),dx$
$\displaystyle \int_{1}^{5}\big((\tfrac{5}{x}+1)-1\big),dx$
$\displaystyle \int_{1}^{5}\big(1-(\tfrac{5}{x}+1)\big),dx$
$\displaystyle \int_{5}^{1}\big(1-(\tfrac{5}{x}+1)\big),dx$
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.