Question

Use the mean value theorem to determine the area of f(x)= 2x^2+3 on the interval [0 , 2]

Answer 1

Answer:

The area of given function is 5.67  unit²

Step-by-step explanation:

Given function f(x) as :

f(x) = 2 x² + 3

The interval in which f(x) lies [ 0 , 2 ]

Let The area for the curve = A

Or ,  A = $\frac{1}{b - a}\int_{a}^{b} f(x) dx$

Or, A = $\frac{1}{2 - 0}\int_{0}^{2} (2x^{2}+3) dx$

Or, A = $\frac{1}{2}(\int_{0}^{2} (2x^{2}) dx + \int_{0}^{2} 3 dx )$

or, A = $\frac{1}{2} [2( \frac{2^{3}-0^{3}}{3})] + \frac{1}{2}[3 (2-0)]$

or, A = $\frac{1}{2}$ ($\frac{16}{3}$ + 6 )

Or, A =  $\frac{1}{2}$ × $\frac{34}{3}$

∴ A = 5.67  unit²

Hence The area of given function is 5.67  unit²   Answer