Area Between Curve and x-axis

Area Between a Curve and the x-axis

Hello JEE aspirants! Welcome to the fascinating world of integral calculus. Today, we're diving into a very important application: finding the area between a curve and the x-axis. This concept is not only fundamental to calculus but also appears frequently in JEE Main, often combined with other topics like coordinate geometry and functions. So, let's get started!

Conceptual Understanding

Imagine you have a curve represented by a function $y = f(x)$. We want to find the area enclosed between this curve, the x-axis, and two vertical lines $x = a$ and $x = b$.

The fundamental idea is that we can approximate this area by dividing it into a large number of very thin rectangles. The width of each rectangle is a small change in $x$, denoted as $dx$, and the height is the value of the function $f(x)$ at that particular $x$. The area of each rectangle is then approximately $f(x) \, dx$.

Summing the areas of all these rectangles gives us an approximation of the total area. As we make the rectangles thinner and thinner (i.e., $dx$ approaches zero), this approximation becomes more and more accurate. This leads us to the concept of the definite integral.

Key Point: The definite integral $\int_a^b f(x) \, dx$ represents the *signed* area between the curve $y = f(x)$ and the x-axis from $x = a$ to $x = b$. Signed area means that area above the x-axis is considered positive, and area below the x-axis is considered negative.

Area Above the x-axis (Positive Area)

If the curve $f(x)$ lies entirely above the x-axis between $x = a$ and $x = b$, then $f(x) \geq 0$ for all $x$ in the interval $[a, b]$. In this case, the definite integral directly gives us the area:

For example, consider the function $f(x) = x^2$ between $x = 1$ and $x = 3$. Since $x^2$ is always positive, the area is simply $\int_1^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_1^3 = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$ square units.

Area Below the x-axis (Absolute Value)

If the curve $f(x)$ lies entirely below the x-axis between $x = a$ and $x = b$, then $f(x) \leq 0$ for all $x$ in the interval $[a, b]$. In this case, the definite integral will give us a negative value. Since area is a non-negative quantity, we need to take the absolute value of the integral:

For instance, consider the function $f(x) = -x^2$ between $x = 1$ and $x = 3$. The integral $\int_1^3 -x^2 \, dx = \left[ -\frac{x^3}{3} \right]_1^3 = -\frac{27}{3} + \frac{1}{3} = -\frac{26}{3}$. Therefore, the area is $\left| -\frac{26}{3} \right| = \frac{26}{3}$ square units.

General Formula: Dealing with Both Positive and Negative Regions

In most JEE problems, the curve will lie both above and below the x-axis within the given interval. In this case, we need to split the integral into regions where $f(x)$ is positive and regions where $f(x)$ is negative, and then take the absolute value of the integral for the negative regions.

This can be expanded as:

If $f(x) \geq 0$ on $[a, c]$ and $f(x) \leq 0$ on $[c, b]$, then:

The point $x = c$ is where the curve intersects the x-axis (i.e., $f(c) = 0$).

Example: Consider $f(x) = x^2 - 4$ between $x = 0$ and $x = 3$. First, find where $f(x) = 0$: $x^2 - 4 = 0 \Rightarrow x = \pm 2$. Since we are considering the interval $[0, 3]$, we have $x = 2$ as the point where the curve crosses the x-axis.

On $[0, 2]$, $f(x) \leq 0$, and on $[2, 3]$, $f(x) \geq 0$. Therefore:

Finding Bounds from the Curve Equation

Sometimes, the limits of integration ($a$ and $b$) are not directly given. Instead, you might be asked to find the area enclosed by the curve and the x-axis. In this case, the bounds are the points where the curve intersects the x-axis. To find these points, solve the equation $f(x) = 0$.

Example: Find the area enclosed by the curve $y = x^2 - 5x + 6$ and the x-axis.

First, find the x-intercepts: $x^2 - 5x + 6 = 0 \Rightarrow (x - 2)(x - 3) = 0 \Rightarrow x = 2, 3$. So, $a = 2$ and $b = 3$.

Since $f(x) \leq 0$ on $[2, 3]$, the area is: