Question

I need to find the area of a shaded region between two curves using definite integrals. The curves are y=x^2+4 and y=x between x=-1 and x=2.

Answer 1

\[Area=\int\limits_{-1}^{2}x^2+4-x=[\frac{ x^3 }{ 3 }-\frac{x^2}{2}+4]_{-1}^{2}=(8/3-4/2-4)-(-1/3-1/2+4)=1.5\]

Answer 2

Correction: \[(8/3-4/2+4)-(-1/3-1/2+4)=9.5\]

Answer 3

\[\large A= \int\limits_{a}^{b}[f(x)-g(x)]dx\]\[\large f(x) = x^2+4\]\[\large g(x) = x\] subtract the function on top from function on bottom to get the area of 1 slice . you're integrating between -1 and 2.\[\large A = \int\limits_{-1}^{2}[(x^2+4)-x]dx\]\[\large A= \int\limits_{-1}^{2}(x^2-x+4)dx\]\[\large A=\frac13x^3-\frac12x^2+4x ]_{-1}^2\]\[\large \quad \quad F(b)-F(a)\]\[\large [\frac13(2)^3 -\frac12(2)^2+4(2)] - [\frac13(-1)^3 -\frac12(-1)^2+4(-1)]\]\[\large \frac{26}{3}-(-\frac{29}{6})=\frac{27}{2}\]

Answer 4

Thank you!!

Answer 5

@shadow_wxh you forgot to integrate the "4" in your equation.

Answer 6

@KyleSeaman Do you understand my method of approach?

Answer 7

Yes I do. I had trouble with the last part, I didn't realize I had to use both 2 and -1 in the final part of the equation.