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Never underestimate the power of stupid people in large groups, George Carlin.
All those who seem stupid are stupid, and so are half of those who don’t seem so, Quevedo.
Antiderivatives are fundamental concepts in calculus, acting as the inverse operation of derivatives.
Given a function $f(x)$, an antiderivative (also known as an indefinite integral) $F$ is the function that can be differentiated to obtain the original function. This means $F' = f$. For example, $x^3 - x + 7$ is the antiderivative of $3x^2 - 1$.
Symbolically, this inverse relationship is written as: $F(x) = \int f(x) \, dx$
The overarching process of finding these antiderivatives is known as integration.
The Fundamental Theorem of Calculus bridges the gap between antiderivatives and geometric area. It states that the definite integral of a function $f$ over an interval is equal to the change in any of its antiderivatives $F$ between the ends of that interval.
Mathematically, if $F'(x) = f(x)$, then: $\boxed{\int_{a}^{b} f(x) \, dx = F(b) - F(a) = F(x) \bigg\vert{}_{a}^{b}}$
The rules of integration allow us to manipulate and simplify complex expressions. Here are the core properties:
Sum of Functions.
The integral of a sum is equal to the sum of the integrals, $\int (f(x) + g(x))dx = \int f(x)dx + \int g(x)dx$.
$\int (8x -5)dx = \int 8x dx -\int 5 dx = \frac{8x^2}{2} + c_1 -5x + c_2 = 4x^2-5x+c$.
Constant Multiple.
You can pull constant multipliers out to the front of the integral, $\int cf(x)dx = c\int f(x)dx$ where c is a constant.
$\int 7\cos(x) \, dx = 7\int \cos(x) \, dx = 7(\sin(x) + c_1) = 7\sin(x) + c_2$
Linearity of Integrals.
Combining the previous two rules, you can split the integral of a linear combination and pull constants out of the integral, $\int (a \cdot f(x) + b \cdot g(x)) \, dx = a \int f(x) \, dx + b \int g(x) \, dx$
$\int (e^{3x} - x^7) \, dx = \int e^{3x} \, dx - \int x^7 \, dx = \frac{e^{3x}}{3} - \frac{x^8}{8} + c$
Zero-Width Interval.
When the upper and lower limits of integration are identical, the integral value is exactly $0$, regardless of the integrand function, representing the area over a single point, $\int_{a}^{a} f(x) \, dx = 0$.
Integrating Definite Integral Backwards.
Reversing the limits of integration flips the sign of the result. Our Fundamental Theorem of Calculus guarantees this works whether $a < b$ or $b < a$.
$\boxed{\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx}$
Proof:
$\int_{a}^{b} f(x) \, dx = F(b) - F(a) = -(F(a) - F(b)) = -\int_{b}^{a} f(x) \, dx$
Definite Integrals on Adjacent Intervals.
Let $a < b < c$. We can break up an integral into smaller, adjacent intervals (or pieces) sharing the same integrand.
Formally, $\boxed{\int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx = \int_{a}^{c} f(x) \, dx}$
$\int_{4}^{7} \sin(3x) \, dx + \int_{-2}^{4} \sin(3x) \, dx = \int_{-2}^{7} \sin(3x) \, dx = -\frac{1}{3}\cos(3x)\bigg\vert{}_{-2}^{7}$
$= \frac{1}{3}(-\cos(21) - (-\cos(-6)))$
Recall that $\cos(-\theta) = \cos(\theta)$: $= \frac{1}{3}(\cos(6) - \cos(21)) \approx 0.5026$
Symmetry (Even and Odd Functions).
Symmetry across the y-axis or origin greatly simplifies integration over bounds like $[-a, a]$.
$\star$ Even Function ($f(-x) = f(x)$).
The area mirrors perfectly, so you can calculate half and double it. $\boxed{\int_{-a}^{a} f(x) \, dx = 2\int_{0}^{a} f(x) \, dx}$
Proof:
$\int_{-a}^{a} f(x)dx$ =[Definite Integrals on Adjacent Intervals] $\int_{-a}^{0} f(x)dx + \int_{0}^{a} f(x)dx = $
Integrating Definite Integral Backwards (splitting at zero): $-\int_{0}^{-a} f(x)dx + \int_{0}^{a} f(x)dx =$
Change of variables for the negative side (u = -x, du = -dx, $u_1 =0, u_2 = a$) yields: $\int_{0}^{a} f(-u)du + \int_{0}^{a} f(x)dx =$
f is an even function, f(-x) = f(x): $\int_{0}^{a} f(u)du + \int_{0}^{a} f(x)dx = 2\int_{0}^{a} f(x)dx$
$\star$ Odd Function ($f(-x) = -f(x)$).
The positive and negative areas cancel each other out entirely. $\boxed{\int_{-a}^{a} f(x) \, dx = 0}$
Proof:
Suppose f is an odd function, $\int_{-a}^{a} f(x)dx =$
Definite Integrals on Adjacent Intervals (splitting at zero): $\int_{-a}^{0} f(x)dx + \int_{0}^{a} f(x)dx =$
Integrating Definite Integral Backwards: $-\int_{0}^{-a} f(x)dx + \int_{0}^{a} f(x)dx$.
Change of variables for the negative side (u = -x, du = -dx, $u_1 =0, u_2 = a$) yields: $\int_{0}^{a} f(-u)du + \int_{0}^{a} f(x)dx =$
f is an odd function, f(-x) = -f(x): $-\int_{0}^{a} f(u)du + \int_{0}^{a} f(x)dx = 0.$
Comparison Theorem.
If one function is always greater than or equal to another, its integral will be as well.
$\star$ If $f(x) \ge 0$ for all $x \in [a, b]$, then $\int_{a}^{b} f(x) \, dx \ge 0$.
$\star$ If $f(x) \le g(x)$ for all $x \in [a, b]$, then $\int_{a}^{b} f(x) \, dx \le \int_{a}^{b} g(x) \, dx$.
Physics interpretation: If I am traveling faster than you over the same time period, I will obviously cover more distance than you.
$\star$ Bounded limits: If $m \le f(x) \le M$ for all $x \in [a, b]$, then the area is sandwiched between two rectangles: $\boxed{m(b-a) \le \int_{a}^{b} f(x) \, dx \le M(b-a)}$
Example 1: Consider $f(x) = \sqrt{1+x^2}$ and $g(x) = \sqrt{1+x}$ over the interval $[0, 1]$. Initially, when graphed on any plotting software, f(x) appears to be above g(x). We need to zoom in to see that g(x) is above f(x) on the interval [0, 1] (Figure 2) ⇒ $\int_{0}^{1} g(x)dx ≥ \int_{0}^{1} f(x)dx$
Example 2: Since $e^x \ge 1$ for all $x \ge 0$ (the exponential function is strictly increasing):
$\int_{0}^{b} e^x \, dx \ge \int_{0}^{b} 1 \, dx$
Evaluating both sides gives $e^b - 1 \ge b$, which rearranges to $e^b \ge 1 + b$ where $b \ge 0$. 🚀
Taking this further, because $e^x \ge 1 + x$ for $x \ge 0$:
$\int_{0}^{b} e^x \, dx \ge \int_{0}^{b} (1 + x) \, dx$
$e^b - 1 \ge \left( x + \frac{x^2}{2} \right)\bigg\vert{}_{0}^{b} = b + \frac{b^2}{2}$
Conclusion: $e^b \ge 1 + b + \frac{b^2}{2}$ where $b \ge 0$.
Integration by substitution or change of variables.
It is a technique used to simplify integrals by introducing a new variable. It is particularly useful when the integrand contains a complicated expression or when the substitution helps to reveal an easier form to calculate the integral, $\int_{x_1}^{x_2} g(u(x)) \cdot u'(x) \, dx = \int_{u_1}^{u_2} g(u) \, du$ where $u = u(x)$, $du = u'(x) \, dx$, and the bounds shift to $u_1 = u(x_1)$ and $u_2 = u(x_2)$. This reveals an easier form to calculate, provided $u'(x)$ doesn’t change signs across the interval.
Evaluate $\int_{1}^{2} (x^3 + 2)^5 x^2 \, dx$
Choose substitution.
Let $u = x^3 + 2$
Find differential and update limits.
$du = 3x^2 \, dx \implies x^2 \, dx = \frac{1}{3} \, du$
$\star$ Lower limit: $u_1 = 1^3 + 2 = 3$
$\star$ Upper limit: $u_2 = 2^3 + 2 = 10$
Rewrite and evaluate.
$\int_{3}^{10} u^5 \left(\frac{1}{3}\right) du = \frac{1}{3} \left[ \frac{u^6}{6} \right]_{3}^{10}= \frac{1}{18} u^6 \bigg\vert{}_{3}^{10} = \frac{1}{18}(10^6 - 3^6) = \frac{1}{18}(1,000,000 - 729) \approx 55,515.056$
Evaluate $\int \frac{1}{1+\sqrt{x}} \, dx$
Choose substitution.
Let $u = \sqrt{x}$
Find differential.
$\frac{du}{dx} = \frac{1}{2\sqrt{x}} \implies dx = 2\sqrt{x} \, du$
Since $u = \sqrt{x}$, we can write this as $dx = 2u \, du$.
Rewrite and evaluate.
$\int \frac{2u \, du}{1+u} = 2 \int \frac{u}{1+u} \, du$
To integrate this, use a clever algebraic trick: add and subtract 1 in the numerator.
$2 \int \frac{u + 1 - 1}{1+u} \, du = 2 \left( \int \frac{u+1}{1+u} \, du + \int \frac{-1}{1+u} \, du \right)= 2 \left( \int 1 \, du - \int \frac{1}{1+u} \, du \right) = 2u - 2\ln\vert{}1+u\vert{} + C$
Finally, substitute $x$ back in: $= 2\sqrt{x} - 2\ln\vert{}1+\sqrt{x}\vert{} + C$
Understand the problem (Figure 1).
To find the area of a circle with radius $R$, we can integrate the function of a semi-circle $y = \sqrt{R^2 - x^2}$ from $0$ to $R$ (which gives one quarter of the circle) and multiply by 4.
$\text{Area} = 4 \int_{0}^{R} \sqrt{R^2-x^2} \, dx$
Choose a trigonometric substitution.
Let $x = R\cos(\theta)$
Find differential and update limits.
$dx = -R\sin(\theta) \, d\theta$
$\star$ Lower limit: $0 = R\cos(\theta_1) \implies \cos(\theta_1) = 0 \implies \theta_1 = \frac{\pi}{2}$
$\star$ Upper limit: $R = R\cos(\theta_2) \implies \cos(\theta_2) = 1 \implies \theta_2 = 0$
Rewrite and evaluate.
$4 \int_{\frac{\pi}{2}}^{0} \sqrt{R^2 - R^2\cos^2(\theta)} (-R\sin(\theta)) \, d\theta$
Factor out $R^2$ and apply the Pythagorean identity $1 - \cos^2(\theta) = \sin^2(\theta)$: $= 4 \int_{\frac{\pi}{2}}^{0} R\sqrt{\sin^2(\theta)} (-R\sin(\theta)) \, d\theta = 4 \int_{\frac{\pi}{2}}^{0} R\sin(\theta) (-R\sin(\theta)) \, d\theta= 4R^2 \int_{0}^{\frac{\pi}{2}} \sin^2(\theta) \, d\theta$
Use the half-angle identity $\sin^2(\theta) = \frac{1-\cos(2\theta)}{2}$: $= 4R^2 \int_{0}^{\frac{\pi}{2}} \frac{1-\cos(2\theta)}{2} \, d\theta = 2R^2 \left[ \int_{0}^{\frac{\pi}{2}} 1 \, d\theta - \int_{0}^{\frac{\pi}{2}} \cos(2\theta) \, d\theta \right] = 2R^2 \left[ \theta - \frac{\sin(2\theta)}{2} \right]_{0}^{\frac{\pi}{2}} = 2R^2 \left( \frac{\pi}{2} - \frac{\sin(\pi)}{2} \right) - 2R^2 \left( 0 - \frac{\sin(0)}{2} \right) = 2R^2 \left( \frac{\pi}{2} - 0 \right) = \pi R^2$
Substitution has rules. If the derivative of your substitution ($u'$) changes signs within your interval of integration, direct substitution of the limits will give you an incorrect answer.
The Flawed Approach: Evaluate $\int_{-1}^{1} x^2 \, dx$
Let $u = x^2$. Then $du = 2x \, dx$, meaning $dx = \frac{du}{2x}$.
If we blindly substitute, we get: $\int_{1}^{1} \frac{\sqrt{u}}{2} \, du = 0$
This is a contradiction. The graph of $y = x^2$ is a positive parabola, so the area under it from $-1$ to $1$ must be strictly greater than zero.
Why it failed: The derivative $u' = 2x$ changes signs at $x = 0$ (it is negative when $x < 0$ and positive when $x > 0$). The function $u = x^2$ is not one-to-one on the interval $[-1, 1]$. To use substitution here, you would have to split the integral at $x=0$.
The Correct Approach (Power Rule): It is much simpler to avoid substitution entirely and use the standard power rule: $\int_{-1}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{1} = \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3} \approx 0.667$
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