Answer to Question #250323 in Calculus for Samkelo

Question

Calculate the mean value of f(x)=3x²+4x+1 between x = −1 and x = 2 

Answer 1

By the Mean Value Theorem for Integrals

$$f_{ave}=\dfrac{1}{2-(-1)}\displaystyle\int_{-1}^{2}(3x^2+4x+1)dx$$

$$=\dfrac{1}{3}[x^3+2x^2+x]\begin{matrix} 2 \\ -1 \end{matrix}$$

$$=\dfrac{1}{3}(8+8+2+1-2+1)=6$$

$$f_{ave}=6$$