Area Between Two Curves Calculator

what is Area between Curves?

In mathematics, the place among curves may be calculated with the difference between the definite vital of two factors or expressions. In -dimensional geometry, the place can explicit with the region covers by using the 2 one-of-a-kind curves. There are two capabilities required to calculate the region, f(x) and g(x) and the crucial limits from a to b wherein b ought to be greater than \(a, b>a\) of the expression. \(Y = f (x) \text{ where x} = a, x = b\) \(Y = f (x)\) between limits of a and b

Method for location between Curves:

we can locate the areas among curves with the aid of using its widespread method if we have distinct curves $$m = f (x) & m = g (x)$$ Where \(f (x) \text{ greater than} g (x)\) So the area bounded by two lines\( x = a \text{ and} x = b\) is $$A = ∫ab[f (x) – g (x)] dx$$ So, the region between curves calculator computes the location wherein two curves intersect every different by means of the use of this fashionable method. however, an on-line integral Calculator lets in you to evaluate the integrals of the capabilities with appreciate to the variable worried.

Example:

Find the area between two curves \( x^2 + 3y - x = 0 \) and the straight line \( x = 2y \).

Solution: The given equations are: 1. \( x^2 + 3y - x = 0 \) 2. \( x = 2y \) **Step 1: Rewrite the first equation:** From \( x^2 + 3y - x = 0 \), solve for \( y \): $$ x^2 + 3y - x = 0 $$ Add \( x \) and subtract \( x^2 \) from both sides: $$ x^2 + 3y - x + x - x^2 = x - x^2 $$ $$ 3y = x - x^2 $$ Divide by 3: $$ y = \frac{x - x^2}{3} $$ **Step 2: Find the points of intersection:** Substitute \( x = 2y \) into the first equation: $$ x^2 + 3 \cdot \frac{x}{2} - x = 0 $$ Simplify: $$ x^2 + \frac{3x}{2} - x = 0 $$ $$ x^2 - \frac{x}{2} = 0 $$ Factorize: $$ x(x - \frac{1}{2}) = 0 $$ Thus, \( x = 0 \) or \( x = \frac{1}{2} \). **Step 3: Correlate \( x \) values to find \( y \):** Using \( x = 2y \): - For \( x = 0 \), \( y = 0 \). - For \( x = \frac{1}{2} \), \( y = \frac{1}{4} \). The points of intersection are \( P(0, 0) \) and \( Q(\frac{1}{2}, \frac{1}{4}) \). **Step 4: Set up the integral for the area:** The curve on the top is \( f(x) = \frac{x}{2} \), and the curve on the bottom is \( g(x) = \frac{x - x^2}{3} \). The area is: $$ A = \int_{0}^{1/2} \left[ f(x) - g(x) \right] dx $$ Substitute the functions: $$ A = \int_{0}^{1/2} \left[ \frac{x}{2} - \frac{x - x^2}{3} \right] dx $$ Simplify: $$ A = \int_{0}^{1/2} \left[ \frac{3x}{6} - \frac{2x - 2x^2}{6} \right] dx $$ $$ A = \int_{0}^{1/2} \left[ \frac{x + 2x^2}{6} \right] dx $$ **Step 5: Solve the integral:** $$ A = \frac{1}{6} \int_{0}^{1/2} \left[ x + 2x^2 \right] dx $$ $$ A = \frac{1}{6} \left[ \frac{x^2}{2} + \frac{2x^3}{3} \right]_{0}^{1/2} $$ Evaluate at \( x = 1/2 \): $$ A = \frac{1}{6} \left[ \frac{(1/2)^2}{2} + \frac{2(1/2)^3}{3} \right] $$ $$ A = \frac{1}{6} \left[ \frac{1}{8} + \frac{1}{12} \right] $$ Find the common denominator: $$ A = \frac{1}{6} \cdot \frac{3 + 2}{24} $$ $$ A = \frac{1}{6} \cdot \frac{5}{24} $$ $$ A = \frac{5}{144} $$ The area is: $$ A = \frac{5}{144} $$

FAQs

Can the location among two Curves be negative or not?

If both the curves lie on the x-axis, so the areas among curves can be bad (-). however, the signed price is the very last answer.

Why we use simplest specific necessary for finding the area Bounded with the aid of Curves?

The location between the curves calculator finds the location by way of distinctive functions simplest indefinite integrals due to the fact indefinite simply suggests the family of various features in addition to use to locate the region between two curves that integrate the difference of the expressions.