How do you find the area between #y=1/2x^3+2, y=x+1, x=0, x=2#?

Question

Scarlett Allison · Accepted Answer

graph{(1/2x^3+2-y)(x+1-y)(x-2)x=0 [-6.67, 9.14, -0.72, 7.18]}

#A=2# First of all #1/2x^3+2>x+1, AA0<=x<=2# To explain this call #f(x)=1/2x^3-x+1# Then #f^'(x)=3/2x^2-1# so #f^'(x)=0# only when #x=+-sqrt(2/3)# and #f^('')(x)=3x>0, AA 0 < x <=2# So f has a local minimum in #x=sqrt(2/3)# and this minimum is #f(sqrt(2/3))=1-2/3sqrt(2/3)>0# Then the area is given by #A=int_0^2(1/2x^3-x+1)dx=[x^4/8-x^2/2+x]_0^2=16/8-4/2+2=2#

Charles Anctil · Answer

undefined To find the area between the curves $y = \frac{1}{2}x^3 + 2$, $y = x + 1$, $x = 0$, and $x = 2$, follow these steps:

1. Find the points of intersection between the curves $y = \frac{1}{2}x^3 + 2$ and $y = x + 1$.
2. Set $y = \frac{1}{2}x^3 + 2$ equal to $y = x + 1$ and solve for $x$.
3. Integrate the difference between the curves from the lower limit to the upper limit.

First, solve for the points of intersection by setting the equations equal to each other:

$$\frac{1}{2}x^3 + 2 = x + 1$$

Solve for $x$:

$$\frac{1}{2}x^3 - x + 1 = 0$$

This equation doesn't have a simple solution, so you may need to use numerical methods or graphing software to find the points of intersection.

Once you find the points of intersection, denote them as $x_1$ and $x_2$.

Finally, integrate the difference between the curves:

$$A = \int_{x_1}^{x_2} (\text{top curve} - \text{bottom curve}) \, dx$$

In this case:

$$A = \int_{x_1}^{x_2} \left((x + 1) - \left(\frac{1}{2}x^3 + 2\right)\right) \, dx$$

Calculate this integral from $x = x_1$ to $x = x_2$ to find the area between the curves.