Area Under a Curve
A carpenter in Santiago de los Caballeros designs a parabolic arch for the entrance of a workshop and needs to know how much sheet metal that shape will cover. Triangles, trapezoids, and circles have formulas, but a curved arch does not. What is, for example, the area of the region between the curve $f(x)=x^2$, the $OX$ axis, and the lines $x=0$ and $x=2$? The definite integral was invented to solve this problem.
Approximating the Area with Rectangles
If you don't know how to find the area of a curved shape, approximate it with shapes whose area you do know how to find. Divide the interval $[a,b]$ into narrow strips of width $\Delta x$ and, on each strip, build a rectangle whose height is the value of $f(x)$ on that strip.
The more rectangles you use (and the narrower they are), the less is "left over" or "missing" between the staircase and the curve, and the sum of their areas gets closer to the true area.
When $f$ is constant, the rectangles coincide with the region itself: the area is exact from the very first rectangle.
An example from physics: the distance traveled is the area under the graph of velocity versus time. An electric scooter rider going at constant speed travels $\int_0^{0.5} 12\,dt$ km in half an hour.
In physics, the distance travelled by an object can be calculated as the area under the curve in a velocity against time graph. If an e-scooter moves at a constant…
In physics, the distance travelled by an object can be calculated as the area under the curve in a velocity against time graph. If an e-scooter moves at a constant velocity of $v(t) = 12$ km/h during the first half hour ($t = 0.5$ h), set up and calculate the definite integral that represents the distance travelled in that time.
The Definite Integral: Notation and Meaning
When the number of rectangles grows without bound ($n\to\infty$ and $\Delta x\to 0$), the sum of their areas settles on a number. That number is the definite integral of $f$ between $a$ and $b$.
When $f$ is always positive on $[a,b]$, $\displaystyle\int_a^b f(x)\,dx$ is exactly the area between the curve and the $OX$ axis. If $f$ is a simple function, you can check that the integral matches the area found by geometry.
To calculate with Barrow's rule, find an antiderivative $F$ of the integrand and evaluate $F(b)-F(a)$.
Consider the linear function $f(x) = \frac{1}{2}x + 1$. You are asked to calculate the area of the shaded region in the graph, bounded by the function, the $OX$-axis,…
Consider the linear function $f(x) = \frac{1}{2}x + 1$. You are asked to calculate the area of the shaded region in the graph, bounded by the function, the $OX$-axis, and the vertical lines $x = 0$ and $x = 4$. To do this:
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Calculate the value of the area using elementary geometric formulae (area of a trapezium).
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Calculate the definite integral $\int_{0}^{4} (\frac{1}{2}x + 1) \, dx$ using Barrow's rule and check that both results coincide.
Calculate the value of the definite integral of the polynomial function $f(x) = 3x^2 - 2x + 5$ in the interval $[1, 2]$. That is, obtain the result of: $$\int_{1}^{2}…
Calculate the value of the definite integral of the polynomial function $f(x) = 3x^2 - 2x + 5$ in the interval $[1, 2]$. That is, obtain the result of:
$$\int_{1}^{2} (3x^2 - 2x + 5) \, dx$$Find the value of a positive parameter $k$, given that the area of the region bounded by the parabola $f(x) = kx^2$ and the $OX$-axis between $x = 0$ and $x = 3$ is…
Find the value of a positive parameter $k$, given that the area of the region bounded by the parabola $f(x) = kx^2$ and the $OX$-axis between $x = 0$ and $x = 3$ is exactly $18$ square units.
The Sign of the Integral: Areas Below the Axis
A Riemann sum does not distinguish "above" from "below": if $f(x)<0$, each rectangle has a negative height and enters the sum with a negative sign.
To find the area between a parabola and the $OX$ axis, first find where it crosses the axis (the roots) and use them as the limits.
A typical case: a car wash tunnel with top edge $f(x)=-x^2+4$ and supports at $x=-2$ and $x=2$. Since $f\ge 0$ on $[-2,2]$, the area is simply the integral.
Find the area of the plane region bounded by the graph of the function $f(x) = x^2 - 9$ and the x-axis ($OX$).
Find the area of the plane region bounded by the graph of the function $f(x) = x^2 - 9$ and the x-axis ($OX$).
A car wash tunnel has a metallic structure whose upper part follows the shape of the parabola $f(x) = -x^2 + 4$. Calculate the area of the flat (front) surface bounded…
A car wash tunnel has a metallic structure whose upper part follows the shape of the parabola $f(x) = -x^2 + 4$. Calculate the area of the flat (front) surface bounded by the parabolic structure and the ground ($OX$-axis), knowing that the support points on the ground are $x = -2$ and $x = 2$.
Calculate the area of the plane region bounded by the graph of the function $f(x) = -x^2 + 6x - 5$ and the x-axis $OX$. Sketch the region before performing the…
Calculate the area of the plane region bounded by the graph of the function $f(x) = -x^2 + 6x - 5$ and the x-axis $OX$. Sketch the region before performing the calculations to identify the x-intercepts.
You can do it on paper and send a photo with your answer.
An architect is designing the entrance of a tunnel whose front cross-section can be modeled by the function $f(x) = 4 - \frac{1}{4}x^2$, where $x$ and $y$ are expressed…
An architect is designing the entrance of a tunnel whose front cross-section can be modeled by the function $f(x) = 4 - \frac{1}{4}x^2$, where $x$ and $y$ are expressed in meters. The floor of the tunnel coincides with the $OX$-axis. Calculate the cross-sectional area of the tunnel entrance.
Functions That Change Sign: You Have to Split the Integral
If $f$ is positive on one part of $[a,b]$ and negative on another, the integral mixes positive and negative areas and does not give the total area (the parts can cancel). Locate the points where $f$ crosses the $OX$ axis and split the integral into as many pieces as there are sign changes.
When the function has an absolute value, such as $f(x)=|x^2-1|$, write it piecewise: $x^2-1$ where $x^2-1\ge0$ and $-(x^2-1)$ where it is negative. That way the integral is split at the points where the expression inside is zero.
A useful property for exercises with data: if $\int_1^4 f=7$ and $\int_4^6 f=-3$, then $\int_1^6 f=7+(-3)=4$. Also, reversing the limits changes the sign: $\int_4^1 2f(x)\,dx=-2\cdot 7=-14$.
Calculate the total area of the region bounded by the line $f(x) = x - 2$, the $OX$-axis, and the vertical lines $x = 0$ and $x = 3$. Note: Notice that the function…
Calculate the total area of the region bounded by the line $f(x) = x - 2$, the $OX$-axis, and the vertical lines $x = 0$ and $x = 3$.
Note: Notice that the function crosses the $OX$-axis at a point inside the given interval, so you will need to split the region into two sub-areas.
Use the properties of the definite integral to solve the following problem: Given that $\int_{1}^{4} f(x) \, dx = 7$ and that $\int_{4}^{6} f(x) \, dx = -3$, calculate,…
Use the properties of the definite integral to solve the following problem:
Given that $\int_{1}^{4} f(x) \, dx = 7$ and that $\int_{4}^{6} f(x) \, dx = -3$, calculate, with justification, the value of:
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$\int_{1}^{6} f(x) \, dx$
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$\int_{4}^{1} 2 \times f(x) \, dx$
Determine the area of the region bounded by the curve $f(x) = x^3 - 4x$ and the $OX$-axis in the interval $[-1, 2]$. Note that the function may change sign within the…
Determine the area of the region bounded by the curve $f(x) = x^3 - 4x$ and the $OX$-axis in the interval $[-1, 2]$. Note that the function may change sign within the interval, so you will need to split the integral if necessary.
Find the area of the region bounded by the function $f(x) = |x^2 - 1|$ and the x-axis between $x = 0$ and $x = 2$. Remember that the definition of modulus implies that…
Find the area of the region bounded by the function $f(x) = |x^2 - 1|$ and the x-axis between $x = 0$ and $x = 2$. Remember that the definition of modulus implies that the function is always non-negative, but its algebraic expression changes at the points where the argument becomes zero.
Area Between Curves and More Complex Regions
Many regions do not rest on the $OX$ axis, but are enclosed between two graphs, or between a curve and a line (tangent, normal, horizontal). The idea is the same: the area is the integral of "the top one minus the bottom one."
Steps for setting up a region:
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Draw the curves (even a rough sketch).
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Find the intersection points by setting the functions equal: these will be the limits of integration.
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Decide which function is on top in each stretch. If they cross inside the interval, split the integral.
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Integrate "top minus bottom" and evaluate with Barrow's rule.
Some problems ask for a parameter: set up the area as a function of $k$ and solve the equation that results from setting it equal to the given value.
Consider the function $f(x) = x^2 - 2x + 2$. Calculate the area of the region bounded by the graph of $f(x)$, the $x$-axis, the $y$-axis, and the tangent line to the…
Consider the function $f(x) = x^2 - 2x + 2$. Calculate the area of the region bounded by the graph of $f(x)$, the $x$-axis, the $y$-axis, and the tangent line to the curve at the point with abscissa $x = 3$.
A graphic design company wants to create a logo that includes the shape of a "wave". This shape is defined by the function $f(x) = \sin(x) + 1$ in the interval…
A graphic design company wants to create a logo that includes the shape of a "wave". This shape is defined by the function $f(x) = \sin(x) + 1$ in the interval $[0, \pi]$. Calculate the total area of the figure bounded by the curve, the $OX$-axis, and the vertical lines $x = 0$ and $x = \pi$.
Calculate the area of the region bounded by the curve $f(x) = \sqrt{x}$, the $OX$-axis, and the line $y = -x + 6$ in the first quadrant. Hint: first identify the point…
Calculate the area of the region bounded by the curve $f(x) = \sqrt{x}$, the $OX$-axis, and the line $y = -x + 6$ in the first quadrant. Hint: first identify the point of intersection between the curve and the line to determine the limits of integration.
An architect is designing the side profile of a new sports hall. The main structure follows the curve of the function $f(x) = -x^2 + 6x$. To ensure stability, a metal…
An architect is designing the side profile of a new sports hall. The main structure follows the curve of the function $f(x) = -x^2 + 6x$. To ensure stability, a metal reinforcement must be placed along the tangent line to the curve at the point with x-coordinate $x = 4$.
Calculate the area of the plane region bounded by the graph of the function $f(x)$, the mentioned tangent line, and the y-axis ($Eje\ OY$).
Consider the functions $f(x) = e^x$ and $g(x) = e^{-x}$. Let $R$ be the region bounded by the graphs of both functions and the vertical line $x = k$ (where $k > 0$).…
Consider the functions $f(x) = e^x$ and $g(x) = e^{-x}$. Let $R$ be the region bounded by the graphs of both functions and the vertical line $x = k$ (where $k > 0$).
Determine the value of the positive parameter $k$ such that the area of this region is exactly $4$ square units.
Consider the function $f(x) = |x^2 - 4|$. We want to study the region of the plane "trapped" between this curve and the horizontal line $y = 5$. Calculate the area of…
Consider the function $f(x) = |x^2 - 4|$. We want to study the region of the plane "trapped" between this curve and the horizontal line $y = 5$.
Calculate the area of the region bounded above by the line $y = 5$ and below by the graph of the function $f(x)$. Note: Take into account the points of intersection and the definition of the modulus function to split the integral if necessary.
Let the function be $f(x) = \ln(x)$. You are asked to: 1. Find the equation of the normal line to the graph of $f(x)$ at the point $P(e, 1)$. 2. Calculate the area of…
Let the function be $f(x) = \ln(x)$. You are asked to:
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Find the equation of the normal line to the graph of $f(x)$ at the point $P(e, 1)$.
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Calculate the area of the plane region bounded by the curve $y = \ln(x)$, the normal line found in the previous section, and the x-axis ($Eje\ OX$).
A water tank has a cross-section whose base is defined by the $OX$-axis and its upper part by the piecewise function:…
A water tank has a cross-section whose base is defined by the $OX$-axis and its upper part by the piecewise function:
$f(x) = \begin{cases} \sqrt{x+4} & \text{if } -4 \leq x < 0 \\ 2 \cos\left(\frac{\pi x}{4}\right) & \text{if } 0 \leq x \leq 2 \end{cases}$
Calculate the total cross-sectional area of the tank by computing the definite integral in each interval.
Determine the value of the constant $a \in \mathbb{R}$ with $a > 1$, knowing that the area of the region bounded by the curve $y = \frac{1}{x^2}$, the $x$-axis, and the…
Determine the value of the constant $a \in \mathbb{R}$ with $a > 1$, knowing that the area of the region bounded by the curve $y = \frac{1}{x^2}$, the $x$-axis, and the vertical lines $x = 1$ and $x = a$ is equal to twice the area of the region bounded by the same curve $y = \frac{1}{x^2}$, the $x$-axis, and the lines $x = a$ and $x = a^2$.
Find the area of the region bounded by the function $f(x) = x^3 - 3x^2 + 2x$ and the horizontal line passing through the local maximum of this function. Hint: You must…
Find the area of the region bounded by the function $f(x) = x^3 - 3x^2 + 2x$ and the horizontal line passing through the local maximum of this function.
Hint: You must first locate the local maximum using the derivative, find the y-coordinate at that point, and then determine the points of intersection between the curve and the resulting line.
Let the region $A$ be the region in the first quadrant bounded by the parabola $y = x^2$ and the line $y = 9$. There is a vertical line with equation $x = c$ (with…
Let the region $A$ be the region in the first quadrant bounded by the parabola $y = x^2$ and the line $y = 9$. There is a vertical line with equation $x = c$ (with $0 < c < 3$) that divides the region $A$ into two subregions of identical area.
Calculate the exact value of the constant $c$.
Back to the beginning: the exact area under $f(x)=x^2$ between $x=0$ and $x=2$ is $\displaystyle\int_0^2 x^2\,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}\ \text{u}^2$, and the carpenter's arch is calculated the same way.