๐ What are Definite Integrals?
A definite integral is a way to find the area under a curve between two specific points. It's a fundamental concept in calculus and has numerous applications in various fields. The process involves finding the antiderivative of a function and evaluating it at the limits of integration.
๐ History and Background
The development of integral calculus is credited to both Isaac Newton and Gottfried Wilhelm Leibniz in the late 17th century. They independently developed the fundamental theorem of calculus, which connects differentiation and integration. Riemann later formalized the definition of the definite integral.
๐ Key Principles of Definite Integrals
- โ Additivity:
If $a < c < b$, then $\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx$.
- ๐ Reversal of Limits:
$\int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx$.
- ๐ Linearity:
$\int_{a}^{b} [cf(x) + dg(x)] dx = c\int_{a}^{b} f(x) dx + d\int_{a}^{b} g(x) dx$, where $c$ and $d$ are constants.
- ๐ Fundamental Theorem of Calculus:
If $F(x)$ is an antiderivative of $f(x)$, then $\int_{a}^{b} f(x) dx = F(b) - F(a)$.
๐ท Example 1: Calculating Distance Traveled
Suppose a car's velocity is given by $v(t) = 3t^2 + 2t$ (in meters per second). Find the distance traveled between $t = 1$ and $t = 3$ seconds.
The distance is the definite integral of the velocity function:
$\int_{1}^{3} (3t^2 + 2t) dt$
First, find the antiderivative:
$F(t) = t^3 + t^2$
Now, evaluate $F(3) - F(1)$:
$F(3) = (3)^3 + (3)^2 = 27 + 9 = 36$
$F(1) = (1)^3 + (1)^2 = 1 + 1 = 2$
So, the distance traveled is $36 - 2 = 34$ meters.
๐ง Example 2: Finding the Area of a Region
Calculate the area of the region bounded by the curve $y = x^2$, the x-axis, and the lines $x = 0$ and $x = 2$.
The area is given by the definite integral:
$\int_{0}^{2} x^2 dx$
The antiderivative of $x^2$ is $\frac{1}{3}x^3$.
Evaluate the antiderivative at the limits:
$\frac{1}{3}(2)^3 - \frac{1}{3}(0)^3 = \frac{8}{3} - 0 = \frac{8}{3}$
So, the area of the region is $\frac{8}{3}$ square units.
โ๏ธ Example 3: Determining Work Done
A force is given by $F(x) = 4x - x^2$ (in Newtons). Find the work done in moving an object from $x = 1$ to $x = 3$ meters.
The work done is the definite integral of the force function:
$\int_{1}^{3} (4x - x^2) dx$
First, find the antiderivative:
$W(x) = 2x^2 - \frac{1}{3}x^3$
Now, evaluate $W(3) - W(1)$:
$W(3) = 2(3)^2 - \frac{1}{3}(3)^3 = 18 - 9 = 9$
$W(1) = 2(1)^2 - \frac{1}{3}(1)^3 = 2 - \frac{1}{3} = \frac{5}{3}$
So, the work done is $9 - \frac{5}{3} = \frac{22}{3}$ Joules.
๐ Example 4: Calculating Average Value of a Function
Find the average value of the function $f(x) = \sin(x)$ on the interval $[0, \pi]$.
The average value is given by:
$\frac{1}{\pi - 0} \int_{0}^{\pi} \sin(x) dx$
The antiderivative of $\sin(x)$ is $-\cos(x)$.
Evaluate the antiderivative at the limits:
$\frac{1}{\pi} [-\cos(\pi) - (-\cos(0))] = \frac{1}{\pi} [-(-1) - (-1)] = \frac{1}{\pi} [1 + 1] = \frac{2}{\pi}$
So, the average value of the function is $\frac{2}{\pi}$.
๐ก๏ธ Example 5: Modeling Population Growth
Suppose the rate of population growth of a bacteria colony is given by $P'(t) = 200e^{0.1t}$ bacteria per hour. Find the increase in population from $t = 0$ to $t = 5$ hours.
The increase in population is the definite integral of the growth rate:
$\int_{0}^{5} 200e^{0.1t} dt$
The antiderivative of $200e^{0.1t}$ is $2000e^{0.1t}$.
Evaluate the antiderivative at the limits:
$2000e^{0.1(5)} - 2000e^{0.1(0)} = 2000e^{0.5} - 2000e^{0} = 2000(e^{0.5} - 1)$
So, the increase in population is approximately $2000(1.6487 - 1) \approx 1297.4$ bacteria.
๐งช Example 6: Calculating Volume of a Solid of Revolution
Find the volume of the solid formed by revolving the region bounded by $y = \sqrt{x}$, the x-axis, and $x = 4$ about the x-axis.
The volume is given by the definite integral:
$V = \pi \int_{0}^{4} (\sqrt{x})^2 dx = \pi \int_{0}^{4} x dx$
The antiderivative of $x$ is $\frac{1}{2}x^2$.
Evaluate the antiderivative at the limits:
$V = \pi [\frac{1}{2}(4)^2 - \frac{1}{2}(0)^2] = \pi [\frac{1}{2}(16) - 0] = 8\pi$
So, the volume of the solid is $8\pi$ cubic units.
๐ Example 7: Determining Consumer Surplus
The demand function for a product is given by $p(x) = 20 - 0.5x$, where $x$ is the number of units and $p$ is the price. If the market price is $p = 10$, find the consumer surplus.
First, find the quantity demanded at $p = 10$:
$10 = 20 - 0.5x \Rightarrow 0.5x = 10 \Rightarrow x = 20$
The consumer surplus is given by the definite integral:
$CS = \int_{0}^{20} (20 - 0.5x) dx - (10)(20)$
Find the antiderivative of $20 - 0.5x$:
$F(x) = 20x - 0.25x^2$
Evaluate the antiderivative at the limits:
$F(20) - F(0) = [20(20) - 0.25(20)^2] - [0] = 400 - 100 = 300$
So, the consumer surplus is $300 - (10)(20) = 300 - 200 = 100$.
๐ฏ Conclusion
Definite integrals are powerful tools with a wide range of real-world applications. From calculating distances and areas to determining work and modeling population growth, understanding definite integrals and antiderivatives is essential in various fields. By mastering these concepts, you can solve complex problems and gain deeper insights into the world around you.