# Sketch the curve y=2e^(x/2)+2 and find area under for 3<=x<=6

### 1 Answer | Add Yours

You need to find the area of the region bounded by the given curves, hence, you need to evaluate the following definite integral, such that:

`int_3^6 (2*e^(x/2) + 2) dx `

Using the property of linearity of integral, you need to split the integral in two simpler integrals, such that:

`int_3^6 (2*e^(x/2) + 2) dx = int_3^6 (2*e^(x/2)) dx + int_3^6 (2) dx `

You need to solve the integral `int_3^6 (2*e^(x/2))dx` using the next substitution, such that:

`x/2 = u => (dx)/2 = du => dx = 2du `

Changing the limits of integration, yields:

`x = 3 => u = 3/2 x = 6 => u = 6/2 => u = 3 `

Changing the variable yields:

`int_(3/2)^3 2*e^u*(2du) = 4*int_(3/2)^3 e^u du `

`4*int_(3/2)^3 e^u du = 4e^u|_(3/2)^3 `

Using the fundamental theorem of calculus yields:

`4*int_(3/2)^3 e^u du = 4(e^3 - e^(3/2))`

Sketching the graph of the curve `y = (2*e^(x/2) + 2)` , between `x = 3, x = 6` , yields:

**Hence, evaluating the area of the region bounded by the given curves, yields **`int_3^6 (2*e^(x/2) + 2) dx = 4(e^3 - e^(3/2)).`