How to find the area under a curve using definite integrals?

In integral calculus, finding the area under a curve and above the x-axis over a specific interval is a fundamental application of definite integrals. Here's a step-by-step guide on how to do this:

1. Identify the interval: Let's say you want to find the area under the curve of the function $f(x)$ from $a$ to $b$. This interval is $[a, b]$.

2. Set up the integral: The definite integral of $f(x)$ from $a$ to $b$ is denoted as: $\int_a^b f(x) \, dx$

3. Evaluate the integral: To find the numerical value of this integral, you'll need to compute the antiderivative (also called the indefinite integral) of $f(x)$. Let's call this antiderivative $F(x)$. So, you have: $\int f(x) \, dx = F(x) + C$ where $C$ is the constant of integration. Now, evaluate $F(x)$ at the upper limit $b$ and lower limit $a$, and subtract the two results: $\left[ F(x) \right]_a^b = F(b) - F(a)$

4. Find the signed area: The result of this calculation gives you the signed area between the curve and the x-axis over the interval $[a, b]$. If $F(b) - F(a)$ is positive, the area is above the x-axis; if it's negative, the area is below the x-axis.

Here's an example: Suppose you want to find the area under the curve of $f(x) = x^2$ from $0$ to $2$. First, find the antiderivative of $f(x)$: $\int x^2 \, dx = \frac{x^3}{3} + C$ Then, evaluate this at the upper limit $2$ and lower limit $0$: $\left[ \frac{x^3}{3} \right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}$ So, the signed area is $\frac{8}{3}$, which means the area above the x-axis is $\frac{8}{3}$ square units.