Area Under a Curve
In a water tank, a swimming pool, or a car wash tunnel, the shape is almost never a triangle or a trapezoid. What is the area of the region between the curve $f(x)=x^2$, the $OX$ axis, and the lines $x=0$ and $x=2$? There is no geometric formula for that shape. The definite integral was created to solve exactly this problem.
Approximating the area with rectangles
If you don't know how to calculate the exact area of a curved shape, approximate it with shapes you do know. 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 excess or shortfall there is between the steps and the actual curve, and the sum of their areas gets closer to the true area.
If $f$ is constant, the region is already a rectangle: the area is exact with a single rectangle.
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 single number. That number is the definite integral of $f$ between $a$ and $b$.
When $f\ge 0$ on $[a, b]$, the integral is exactly the area between the curve and the $OX$ axis. If $f$ is simple (a line, for example), you can check this with geometry.
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:
-
Calculate the value of the area using elementary geometric formulae (area of a trapezium).
-
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$$Find the value of a positive parameter $k$, given that the area of the region bounded by the parabola $f(x) = kx^2$ and the $OX$-axis between $x = 0$ and $x = 3$ is…
Find the value of a positive parameter $k$, given that the area of the region bounded by the parabola $f(x) = kx^2$ and the $OX$-axis between $x = 0$ and $x = 3$ is exactly $18$ square units.
Antiderivatives for computing the integral
To compute a definite integral without adding up rectangles, you use the Fundamental Theorem of Calculus: if $F$ is an antiderivative of $f$ (that is, $F'(x)=f(x)$), then
$$\int_a^b f(x)\,dx=F(b)-F(a)=\Big[F(x)\Big]_a^b.$$An antiderivative is found by reading the table of derivatives backwards. These are the ones you need in this section:
| $f(x)$ | An antiderivative $F(x)$ |
|---|---|
| $x^n$ (with $n\ne -1$) | $\dfrac{x^{n+1}}{n+1}$ |
| $\sin x$ | $-\cos x$ |
| $\cos x$ | $\sin x$ |
| $e^x$ | $e^x$ |
| $\dfrac{1}{x}$ | $\ln x$ (for $x>0$) |
| $\ln x$ | $x\ln x - x$ |
Why? Just differentiate the right column. For example, $(-\cos x)'=\sin x$ and $(x\ln x-x)'=\ln x+1-1=\ln x$.
If the argument is linear, $kx$, divide by $k$: $\displaystyle\int\cos(kx)\,dx=\frac{\sin(kx)}{k}$ and $\displaystyle\int e^{kx}\,dx=\frac{e^{kx}}{k}$. This is what the chain rule requires when you differentiate back.
Tangent line and normal line
Several problems use a line associated with the curve. The tangent at $x=x_0$ has slope $f'(x_0)$. The normal is perpendicular to it, so its slope is $-\dfrac{1}{f'(x_0)}$ (if $f'(x_0)\ne0$).
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 $x = \pi$.
The sign of the integral: areas below the axis
A Riemann sum doesn't distinguish "above" from "below": if $f(x)<0$, each rectangle has negative height and enters the sum with a negative sign.
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.
Functions that change sign: you have to split the integral
If $f$ is positive on part of $[a, b]$ and negative on another part, $\int_a^b f(x)\,dx$ mixes positive and negative areas and does not give the total area: the parts can cancel out. To find the real area, locate the points where $f$ crosses the $OX$ axis and split the integral into as many pieces as there are sign changes.
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:
-
$\int_{1}^{6} f(x) \, dx$
-
$\int_{4}^{1} 2 \times f(x) \, dx$
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.
Areas of regions: absolute value, piecewise, and between curves
Many problems ask for the area of a region bounded by several curves. The method is always the same: draw a sketch, find the intersection points, and decide which curve is on top in each stretch.
For functions with absolute value or defined piecewise, write each piece without bars and integrate separately, for example $|x^2-1|=x^2-1$ if $|x|\ge1$ and $1-x^2$ if $|x|<1$.
A graphic design company wants to create a logo that includes a curved shape defined by $f(x) = \sqrt{x}$. Calculate the area of the part of the logo enclosed between…
A graphic design company wants to create a logo that includes a curved shape defined by $f(x) = \sqrt{x}$. Calculate the area of the part of the logo enclosed between this curve, the $OX$-axis, and the vertical line $x = 9$.
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.
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.
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 $x = 3$.
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:
-
Find the equation of the normal line to the graph of $f(x)$ at the point $P(e, 1)$.
-
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 area under $f(x)=x^2$ between $x=0$ and $x=2$ is $\int_0^2 x^2\,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}\approx 2.67$ units². With $4$ rectangles the estimate gives $3.75$ and, with more rectangles, it gets closer and closer to $\frac{8}{3}$.