Area between Two Curves Calculator

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**Step 1 :**

**Step 2 :**

**Step 3 :**

Area between two curves is the area under which the two curves are intersected. **Area between Two Curves Calculator** calculates the area under which the two curve intersects.

You can see two default functions with its lower and upper bound given below. When you click on "Calculate Area", the calculator will first find the intersecting points and then find the area bounded by the difference between the two curves.

You can see two default functions with its lower and upper bound given below. When you click on "Calculate Area", the calculator will first find the intersecting points and then find the area bounded by the difference between the two curves.

Given are the two curves y = f(x) and y = g(x). equating both the curves to get the value of interval [a,b] where the given curves intersects.

The area between the two curves bounded by the x axis is given by the formula

A = $\int_{a}^{b}$[f(x) − g(x)]dx where f(x) is upper curve and g(x) is the lower curve.

Simplify the problem by substituting the value of the limit to get the answer.

Find the area bounded by the curve y = 6x + x

^{3}below by y = 5x^{2}.**Step 1 :**Find the intersection points by equalizing the two curves: 6x + x

^{3}= 5x^{2}.

6 + x^{2}= 5x.

x^{2}- 5x + 6 = 0

x^{2}- 3x - 2x + 6 = 0

x(x-3) -2 (x-3) = 0

The two points where the curves intersects are x=2 and x=3.**Step 2 :**The difference between these curves at intersection points are given by:

A = $\int_{2}^{3}$ $[(6x + x^{3}) - 5x^{2}]$ dx

= $\int_{2}^{3}$ $[ x^{3} - 5x^{2} + \frac{x^{2}}{2}]$ dx

= $[\frac{x^{4}}{4} - 5 \frac{x^{3}}{3} + 3 \frac{x^{2}}{2}]^{3}_{2}$

= [$\frac{3^{4}}{4}$ - 5 $\frac{3^{3}}{3}$ + 3 (3)2] - [$\frac{2^{4}}{4}$ - 5 $\frac{2^{3}}{3}$ + 3(2)^{2}].

**Step 3 :**A = [$\frac{81 \times 3 + 54 \times 6 - 135 \times 4 - 16 \times 3 - 24 \times 6 + 40 \times 4 }{12}$]

= - $\frac{5}{12}$.**Answer :**The area between two curves is - $\frac{5}{12}$.

Find the area between the curves y = 2 - x

^{2}and y = x.**Step 1 :**given curves are y = 2 - x

^{2}and y = x.

equating both the curves we get,

- x

^{2}+ x + 2 = 0

- x^{2}+ 2x - x + 2 = 0

-x(x - 2) - (x - 2) = 0

(-x - 1)(x - 2) = 0

x = -1, 2 are the two intersecting points.**Step 2 :**The difference between these curves at intersection points are given by:

A = $\int_{-1}^{2} [(2 - x^{2}) - x]$ dx

= $\left[ 2x - \frac{x^{3}}{3} - \frac{x^{2}}{2} \right]_{1}^{-2}$

= $\left[ 2(1) - \frac{1^{3}}{3} - \frac{1^{2}}{2} \right]_{1}^{-2} - \left[ 2(-2) - \frac{(-2)^{3}}{3} - \frac{(-2)^{2}}{2} \right]$

**Step 3 :**A = $\left[2 - \frac{1}{3} - \frac{1}{2} + 4 - \frac{8}{3} + \frac{4}{2} \right]$

= $\frac{9}{2}$

**Answer :**The Area between two curves is $\frac{9}{2}$.