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He may look like an idiot and talk like an idiot but don’t let that fool you. He really is an idiot, Groucho Marx.
Antiderivatives are the inverse operation of derivatives and form the foundation of integral calculus.
Given a function $f(x)$, an antiderivative (also known as an indefinite integral), denoted as $F(x)$, is a function that can be differentiated to obtain the original function, i.e., $F'(x) = f(x)$; e.g., $3x^2 - 1$ is the derivative of $x^3 - x + 7$, which means $x^3 - x + 7$ is an antiderivative of $3x^2 - 1$.
Symbolically, the process of finding antiderivatives is called integration, and we write it as: $F(x) = \int f(x)dx$
The Fundamental Theorem of Calculus bridges the gap between derivatives and integrals.
It states that the definite integral of a function $f$ over an interval $[a, b]$ is equal to the change in any of its antiderivatives $F$ between the ends of that interval: $\boxed{\int_{a}^{b} f(x)dx = F(b) - F(a) = F(x) \bigg\vert{}_{a}^{b}}$
If we consider $f$ to be a continuous rate of change (meaning $f(x) = F'(x)$), this theorem can be restated as the Net Change Theorem: The definite integral of a rate of change of a function on $[a, b]$ is the total change of the function itself on that interval.
The Second Fundamental Theorem of Calculus. If $f$ is a continuous function and c is a constant, then f has a unique antiderivative A that satisfies $A(c) = 0$, and that antiderivative is given by the rule $A(x) = \int_{c}^{x} f(t)dt$.
In fundamental calculus, the definite integral $\int_{a}^{b} f(x) \, dx$ gives us the signed area between the curve $y = f(x)$ and the x-axis. We can extend this logic to find the area bounded between two different functions.
Suppose we want to determine the area of the region bounded by $y = f(x)$ on the top, $y = g(x)$ on the bottom, and the vertical lines $x = a$ and $x = b$.
Conceptually, imagine slicing the region into infinitely many thin vertical rectangles. The height of each rectangle is the difference between the upper curve and the lower curve at a specific point $x$, which is $(f(x) - g(x))$. The width of each rectangle is an infinitesimally small change in $x$, denoted as $dx$. Since the area of a rectangle is base $\times$ height, the area of one tiny slice is $(f(x) - g(x)) \, dx$ -Figure 1.d.-.
By summing all these infinite slices across the interval $[a, b]$ using a definite integral, we arrive at the general formula: $\text{Area} = \int_{a}^{b} \vert{}f(x) - g(x)\vert{} \, dx$
Note: We use the absolute value to ensure we are calculating a positive physical area, though in practice, we usually just subtract the lower function from the upper function. For simplicity sake, we are also going to assume that $f(x) \gt g(x), \boxed{\text{Area} = \int_{a}^{b} f(x) - g(x) \, dx}$
Let’s find the area of the region enclosed by $y = x^2 + 1$, $y = x$, $x = 0$, and $x = 1$ (Figure 4).
Identify the region bounded by the given curves.
Determine the interval $[a, b]$, the points of intersection. Sometimes this is given directly (like $x=0$ to $x=1$). Other times, the region is fully enclosed by the curves, meaning you must set $f(x) = g(x)$ and solve for $x$ to find where they intersect.
In this example, let’s check for intersections: $x^2 + 1 = x \implies x^2 - x + 1 = 0$
Using the quadratic formula, $x = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2}$. Since the discriminant is negative, the solutions are complex, which means the curves do not intersect in the real plane.
We will just use the given interval: $a = 0$ and $b = 1$.
Determine the upper and lower functions.
Identify which function lies above the other on the interval $[a, b]$. You can do this by plugging in a test value between $a$ and $b$.
Let’s test $x = 0.5$: $f(0.5) = (0.5)^2 + 1 = 1.25$ and $g(0.5) = 0.5$.
Since $1.25 > 0.5$, the parabola $y = x^2 + 1$ is the upper function.
Integrate the difference.
Set up the integral by subtracting the lower function from the upper function, then evaluate.
$\text{Area} = \int_{0}^{1} (x^2 + 1 - x) \, dx = \left[ \frac{x^3}{3} - \frac{x^2}{2} + x \right]_{0}^{1}$
$= \left( \frac{1}{3} - \frac{1}{2} + 1 \right) - (0) = \frac{2}{6} - \frac{3}{6} + \frac{6}{6} = \frac{5}{6} \text{ square units}$
Find intersection points.
Set the equations equal to each other to find where the boundaries meet:
$x^2 - 4x = 2x$. Simplify, $x^2 - 6x = 0 \implies x(x - 6) = 0$
The curves intersect at $x = 0$ and $x = 6$. These are our bounds of integration.
Determine the upper and lower functions.
Testing a point like $x = 1$ reveals that $y = 2(1) = 2$ is higher than $y = 1^2 - 4(1) = -3$.
Therefore, $y = 2x$ is the upper function.
Integrate the difference.
$\text{Area} = \int_{0}^{6} [2x - (x^2 - 4x)] \, dx = \int_{0}^{6} (-x^2 + 6x) \, dx$
$= \left[ -\frac{x^3}{3} + 3x^2 \right]_{0}^{6} = \left( -\frac{216}{3} + 3(36) \right) - 0 = -72 + 108 = 36 \text{ square units}$
Problem: Find the area bounded by $y = x^2$, the y-axis ($x=0$), and the horizontal line $y = 16$ (Figure B).
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There are two ways to solve this problem, illustrating the flexibility of calculus.
Imagine a large rectangle formed by the bounds $x=0$, $x=4$, $y=0$, and $y=16$. The area of this full rectangle is $4 \times 16 = 64$.
The area under the parabola $y = x^2$ inside this rectangle is: $\int_{0}^{4} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{4} = \frac{64}{3}$
To find the bounded area above the parabola and below $y=16$, we simply subtract the area under the curve from the total rectangle: $\text{Area} = 64 - \frac{64}{3} = \frac{192}{3} - \frac{64}{3} = \frac{128}{3} \text{ square units}$
Instead of integrating with respect to $x$, we can integrate with respect to $y$ by slicing the region horizontally.
First, rewrite the function $y = x^2$ in terms of $y$: $x = \sqrt{y}$.
Now, integrate from the bottom bound ($y=0$) to the top bound ($y=16$): $\text{Area} = \int_{0}^{16} \sqrt{y} \, dy = \int_{0}^{16} y^{1/2} \, dy = \left[ \frac{2}{3}y^{3/2} \right]_{0}^{16}$
$= \frac{2}{3}(16^{3/2}) = \frac{2}{3}(4^3) = \frac{2}{3}(64) = \frac{128}{3} \text{ square units}$
Both methods give the exact same result!
Problem: Calculate the area bounded by the curves $y = \pm\sqrt{\frac{1}{2}x}$ and $y = \pm\sqrt{x - 4}$, the green-shaded area (Figure C).
Dealing with $\pm$ square roots makes integrating with respect to $x$ messy. Instead, let’s rewrite the equations as functions of $y$:
$\star \quad y = \pm\sqrt{\frac{1}{2}x} \implies y^2 = \frac{1}{2}x \implies x = 2y^2$
$\star \quad y = \pm\sqrt{x - 4} \implies y^2 = x - 4 \implies x = y^2 + 4$
Now we are using horizontal slices of height $dy$ and width $(x_{\text{right}} - x_{\text{left}})$.
Let’s find the intersection points by setting the equations equal to each other: $2y^2 = y^2 + 4 \implies y^2 = 4 \implies y = \pm 2$
Our limits of integration are from $y = -2$ to $y = 2$. Testing $y = 0$ shows that $x = y^2 + 4$ is the curve further to the right.
Defining the integral. $\text{Area} = \int_{-2}^{2} \left( (y^2 + 4) - 2y^2 \right) \, dy = \int_{-2}^{2} (4 - y^2) \, dy$.
Solve it. $\text{Area} = \left[ 4y - \frac{y^3}{3} \right]_{-2}^{2} = \left( 8 - \frac{8}{3} \right) - \left( -8 + \frac{8}{3} \right)$
Simplify, $\text{Area} = 16 - \frac{16}{3} = \frac{32}{3} \text{ units}^2$
Problem: Find the area between the parabola $x = y^2$ and the line $y = x - 2$.
First, calculate the points of intersection by substituting $x$: $y = (y^2) - 2 \implies y^2 - y - 2 = 0 \implies (y - 2)(y + 1) = 0$
The curves intersect at $y = 2$ (point: $4, 2$) and $y = -1$ (point: $1, -1$).
Here is a comparison of solving this problem using vertical versus horizontal slices:
Using vertical slices is difficult here because the “bottom” boundary changes.
From $x = 0$ to $x = 1$, the bottom boundary is the lower half of the parabola ($y = -\sqrt{x}$). From $x = 1$ to $x = 4$, the bottom boundary is the line ($y = x - 2$), Figure 1.e.
Therefore, we have to split the integral into two parts: $\text{Area} = \int_{0}^{1} \left( \sqrt{x} - (-\sqrt{x}) \right) \, dx + \int_{1}^{4} \left( \sqrt{x} - (x - 2) \right) \, dx$.
$\text{Area} = \int_{0}^{1} 2\sqrt{x} \, dx + \int_{1}^{4} (\sqrt{x} - x + 2) \, dx = \left[ \frac{4}{3}x^{3/2} \right]_{0}^{1} + \left[ \frac{2}{3}x^{3/2} - \frac{1}{2}x^2 + 2x \right]_{1}^{4}$
$\text{Area} = \left( \frac{4}{3} \right) + \left( \left[ \frac{16}{3} - 8 + 8 \right] - \left[ \frac{2}{3} - \frac{1}{2} + 2 \right] \right) = \frac{4}{3} + \frac{16}{3} - \frac{13}{6} = \frac{9}{2} \text{ units}^2$
Because the rightmost curve is always the line ($x = y + 2$) and the leftmost curve is always the parabola ($x = y^2$), we can solve this with a single integral from $y = -1$ to $y = 2$, namely: $\text{Area} = \int_{-1}^{2} \left( (y + 2) - y^2 \right) \, dy = \left[ \frac{y^2}{2} + 2y - \frac{y^3}{3} \right]_{-1}^{2}$, figure 1.f.
$\text{Area} = \left( \frac{4}{2} + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 - \left( -\frac{1}{3} \right) \right) = \left( 6 - \frac{8}{3} \right) - \left( -\frac{3}{2} + \frac{1}{3} \right) = \frac{10}{3} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} \text{ units}^2$
Problem: Compute the area between $f(x) = -x^2 + 4x$ and $g(x) = x^2 - 6x + 5$ strictly over the interval $[0, 1]$ (Figure 3).
Calculate the point of intersection:
Set the functions equal: $-x^2 + 4x = x^2 - 6x + 5 \implies 2x^2 - 10x + 5 = 0$.
Using the quadratic formula: $x = \frac{10 \pm \sqrt{100 - 40}}{4} = \frac{5 \pm \sqrt{15}}{2}$.
Only the smaller root, $x = \frac{5 - \sqrt{15}}{2}$ (approximately $0.56$), falls inside our interval $[0, 1]$. For brevity, let’s call this intersection point $i$.
Determine intervals of interest:
The area must be split into two regions: $[0, i]$ and $[i, 1]$.
On $[0, i]$, testing $x=0$ shows $g(0)=5$ and $f(0)=0$. Thus, $g(x) > f(x)$.
On $[i, 1]$, testing $x=1$ shows $f(1)=3$ and $g(1)=0$. Thus, $f(x) > g(x)$.
Integrate the absolute difference:
Set up the two integrals:
$\int_{0}^{i} \left( (x^2 - 6x + 5) - (-x^2 + 4x) \right) \, dx + \int_{i}^{1} \left( (-x^2 + 4x) - (x^2 - 6x + 5) \right) \, dx$
Simplify, $= \int_{0}^{i} (2x^2 - 10x + 5) \, dx + \int_{i}^{1} (-2x^2 + 10x - 5) \, dx$
Evaluating these bounds yields the exact area: $\frac{-52}{3} + 5\sqrt{15} \text{ units}^2$.
Problem: Calculate the area of the shaded region bounded by $y = -x - 1$, $y = 1 - x^2$, and $y = x - 1$ (Figure A).
Calculate the point of intersection:
First, find where the top parabola $y = 1 - x^2$ intersects the two lines:
With $y = -x - 1$: $1 - x^2 = -x - 1 \implies x^2 - x - 2 = 0 \implies x = 2, x = -1$.
With $y = x - 1$: $1 - x^2 = x - 1 \implies x^2 + x - 2 = 0 \implies x = 1, x = -2$.
Determine intervals of interest:
The bounded region is a triangle-like shape under the parabola. The parabola is the “top” curve the entire time, but the “bottom” boundary changes at the $y$-axis ($x = 0$):
From $x = -1$ to $x = 0$, the bottom curve is $y = -x - 1$.
From $x = 0$ to $x = 1$, the bottom curve is $y = x - 1$.
Using Symmetry, calculate the area:
Because the parabola and the bounding lines are perfectly symmetric across the $y$-axis, we only need to calculate the area of the right half ($x = 0$ to $x = 1$) and double it:
$\text{Area} = 2 \int_{0}^{1} \left( (1 - x^2) - (x - 1) \right) \, dx = 2 \int_{0}^{1} (-x^2 - x + 2) \, dx$
$\text{Area} = 2 \left[ -\frac{x^3}{3} - \frac{x^2}{2} + 2x \right]_{0}^{1}$
$\text{Area} = 2 \left( -\frac{1}{3} - \frac{1}{2} + 2 \right) = 2 \left( -\frac{2}{6} - \frac{3}{6} + \frac{12}{6} \right) = 2 \left( \frac{7}{6} \right) = \frac{7}{3} \text{ units}^2$.
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