Area Between Two Curves

Hey JEE aspirants! Welcome to an exciting journey into the world of integral calculus, where we'll learn to calculate the area nestled between two curves. This topic is not just mathematically elegant, but also crucial for scoring well in JEE Main. Understanding area between curves helps you visualize and solve a variety of problems, making it a must-know concept.

Conceptual Explanation

Imagine you have two functions, $f(x)$ and $g(x)$ , plotted on a graph. The area between these curves over an interval $[a, b]$ is the region enclosed by the curves and the vertical lines $x = a$ and $x = b$ . Intuitively, we're summing up infinitely thin rectangles between the curves.

The core idea is that the height of each rectangle is the difference between the y-values of the two curves at a given x-value, i.e., $|f(x) - g(x)|$ . This difference ensures we always get a positive height, regardless of which curve is on top.

Important Formulas

Formula 1: General Area Between Two Curves

The most general formula for the area between two curves $f(x)$ and $g(x)$ from $x = a$ to $x = b$ is:

Area = \int_{a}^{b} |f(x) - g(x)| \, dx

The absolute value ensures that we are always integrating a positive difference, giving us the area regardless of which function is greater on different intervals.

Formula 2: Area When One Curve is Always Above

If $f(x) \geq g(x)$ for all $x$ in the interval $[a, b]$ , meaning $f(x)$ is always greater than or equal to $g(x)$ on the interval, then the formula simplifies to:

Area = \int_{a}^{b} [f(x) - g(x)] \, dx

Here, $f(x)$ is the upper curve, and $g(x)$ is the lower curve. We're essentially integrating the difference between the two functions.

Example: Consider $f(x) = x^2 + 2$ and $g(x) = x$ on the interval $[0, 2]$ . Clearly, $f(x) \geq g(x)$ on this interval. The area is:

Area = \int_{0}^{2} [(x^2 + 2) - x] \, dx = \int_{0}^{2} (x^2 - x + 2) \, dx

Area = \left[ \frac{x^3}{3} - \frac{x^2}{2} + 2x \right]_{0}^{2} = \frac{8}{3} - 2 + 4 = \frac{14}{3}

Finding Intersection Points

To apply these formulas, you often need to find the intersection points of the curves, as these define the limits of integration ( $a$ and $b$ ).

Set the functions equal: Solve $f(x) = g(x)$ for $x$ .
Solve for x: The solutions are the x-coordinates of the intersection points.

Example: Find the intersection points of $f(x) = x^2$ and $g(x) = 2x$ .

x^2 = 2x \Rightarrow x^2 - 2x = 0 \Rightarrow x(x - 2) = 0

So, $x = 0$ and $x = 2$ are the intersection points.

When Curves Cross Each Other

Sometimes, the curves $f(x)$ and $g(x)$ cross each other within the interval $[a, b]$ . In such cases, you need to split the integral into sub-intervals where one function is consistently above the other.

Find all intersection points within $[a, b]$ .
Divide the interval into sub-intervals based on these points.
Integrate separately over each sub-interval, using the correct upper and lower functions.
Add the absolute values of these integrals to get the total area.

Example: Find the area between $f(x) = x^3$ and $g(x) = x$ from $x = -1$ to $x = 1$ .

First, find intersection points: $x^3 = x \Rightarrow x^3 - x = 0 \Rightarrow x(x^2 - 1) = 0$ . So, $x = -1, 0, 1$ .

Now, split the integral:

Area = \int_{-1}^{0} |x^3 - x| \, dx + \int_{0}^{1} |x^3 - x| \, dx

On $[-1, 0]$ , $x^3 \geq x$ , and on $[0, 1]$ , $x \geq x^3$ .

Area = \int_{-1}^{0} (x^3 - x) \, dx + \int_{0}^{1} (x - x^3) \, dx

Area = \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{-1}^{0} + \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{0}^{1} = \left(0 - \left(\frac{1}{4} - \frac{1}{2}\right)\right) + \left(\frac{1}{2} - \frac{1}{4} - 0\right) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}

Formula 3: Integrating with Respect to y

Sometimes, it's easier to integrate with respect to $y$ . In this case, we express $x$ as a function of $y$ , say $x = f(y)$ and $x = g(y)$ , and the area between the curves from $y = c$ to $y = d$ is:

Area = \int_{c}^{d} |f(y) - g(y)| \, dy

Here, $f(y)$ is the "right" curve and $g(y)$ is the "left" curve. The formula becomes:

Area = \int_{c}^{d} [\text{right curve} - \text{left curve}] \, dy

Example: Find the area between $x = y^2$ and $x = 2 - y^2$ .

First, find intersection points: $y^2 = 2 - y^2 \Rightarrow 2y^2 = 2 \Rightarrow y^2 = 1$ . So, $y = -1$ and $y = 1$ .

On $[-1, 1]$ , $2 - y^2 \geq y^2$ .

Area = \int_{-1}^{1} [(2 - y^2) - y^2] \, dy = \int_{-1}^{1} (2 - 2y^2) \, dy

Area = \left[ 2y - \frac{2y^3}{3} \right]_{-1}^{1} = \left(2 - \frac{2}{3}\right) - \left(-2 + \frac{2}{3}\right) = 4 - \frac{4}{3} = \frac{8}{3}

Tips for Solving Problems

Always draw a sketch of the curves. This helps visualize the region and identify the upper and lower functions.
Be careful with absolute values. Split the integral if the curves cross within the interval.
Consider integrating with respect to $y$ if the curves are easier to express as $x = f(y)$ .
Check your limits of integration. Ensure they are the correct intersection points or boundaries of the region.

Common Mistakes to Avoid

Forgetting the absolute value when the curves cross each other.
Incorrectly identifying the upper and lower functions.
Using wrong limits of integration.
Not splitting the integral when curves cross.
Algebraic errors while solving for intersection points or integrating.

JEE-Specific Tricks

For JEE Main, efficiency is key. Here are some tricks to help you solve problems faster:

Symmetry: If the region is symmetric about the x or y-axis, use this to simplify the integral. For example, if the area is symmetric about the y-axis: $Area = 2 \int_{0}^{b} [f(x) - g(x)] \, dx$
Standard integrals: Memorize standard integrals to save time.
Approximation: Use numerical approximation techniques if you're stuck or need to check your answer quickly.

With these concepts, formulas, tips, and tricks, you're well-equipped to tackle area between curves problems in JEE Main. Keep practicing, and you'll master this important topic!