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 = [tex]\frac{1}{b - a}\int_{a}^{b} f(x) dx[/tex]
Or, A = [tex]\frac{1}{2 - 0}\int_{0}^{2} (2x^{2}+3) dx[/tex]
Or, A = [tex]\frac{1}{2}(\int_{0}^{2} (2x^{2}) dx + \int_{0}^{2} 3 dx )[/tex]
or, A = [tex]\frac{1}{2} [2( \frac{2^{3}-0^{3}}{3})] + \frac{1}{2}[3 (2-0)][/tex]
or, A = [tex]\frac{1}{2}[/tex] ([tex]\frac{16}{3}[/tex] + 6 )
Or, A = [tex]\frac{1}{2}[/tex] × [tex]\frac{34}{3}[/tex]
∴ A = 5.67 unit²
Hence The area of given function is 5.67 unit² Answer
