Answer to Question #250323 in Calculus for Samkelo

Question #250323

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


1
Expert's answer
2021-10-14T11:04:39-0400

By the Mean Value Theorem for Integrals


fave=12(1)12(3x2+4x+1)dxf_{ave}=\dfrac{1}{2-(-1)}\displaystyle\int_{-1}^{2}(3x^2+4x+1)dx

=13[x3+2x2+x]21=\dfrac{1}{3}[x^3+2x^2+x]\begin{matrix} 2 \\ -1 \end{matrix}

=13(8+8+2+12+1)=6=\dfrac{1}{3}(8+8+2+1-2+1)=6

fave=6f_{ave}=6


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