Area between two curves calculator will help to calculate the area of the region covered under the two interdected curves. Area is computed by the definite integral. If y = f(x) and y = g(x) are the two curves on the interval [a,b], such that f(x)>g(x), then the area bounded by the lines x = a and x = b is,

Plug in the two functions with the intervals into the area between two curves calculator and area is calculated.

Area = $\int_{a}^{b}[upper \ curve - lower \ curve]dx$

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

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

Plug in the two functions with the intervals into the area between two curves calculator and area is calculated.

Step 2 : Substitute the intersecting points in the area between the two curve formula given above. Simplify by substituting the limit and get the area bounded.

Below are the problems based on Area Between two curves.

### Solved Examples

**Question 1:**Determine the are bounded by the curve $y = 2x^2 + 10$ and $y = 4x + 16$ .

**Solution:**

Step 1 : Given $f(x) = 2x^2 + 10$

$g(x) = 4x + 16$

$2x^2 + 10 = 4x + 16$

$2x^2 - 4x - 6 = 0$

$2(x+1)(x-3) = 0$

The two intersecting points are x = -1 and x = 3.

Step 2 : Area bounded between the two curve,

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

= $\int_{-1}^{3}[(2x^2 + 10)-(4x + 16)]dx$

= $\int_{-1}^{3}[2x^2 - 4x - 6]dx$

= [2 $\frac{x^{3}}{3}$ - 2 ${x^{2}} - 6x]^{3}_{2}$

= - $\frac{64}{3}$

Therefore, the area bounded between the two curve is - $\frac{64}{3}$.

**Question 2:**Determine the are bounded by the curve $y = x^2$ and $y = \sqrt{x}$ .

**Solution:**

Step 1 :Given $f(x) = \sqrt{x}$

$g(x) = x^2$

$\sqrt{x} = x^2$

$x = 0, 1$

The intersecting points are x = 0, 1.

Step 2 : Area bounded between the two curve,

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

= $\int_{-1}^{3}[(\sqrt{x})-(x^2)]dx$

= [$\frac{2}{3}$$x^{\frac{3}{2}}$-$\frac{1}{3}$$x^{3}]^{1}_{0}$

= $\frac{1}{3}$

Therefore, the area bounded between the two curve is $\frac{1}{3}$.