Question 111411
Find the approximate area of the region bounded by {{{y = (1/2)x^2+2}}}, {{{x = 0}}}, {{{x = 3}}}, and {{{y = 0}}}
First, it would be helpful to see the graph of the equation:
{{{graph(600,400,-5,5,-5,8,(1/2)x^2+2)}}}
Now divide the region into two areas by a vertical line through x = 2 and another vertical line through x = 3.
You can approximate the shape of these two area by trapezoids (lying on their sides).
The first trapezoid {{{b[1] = 2}}}and{{{b[2] = 4}}}
How did I get these?
Just substitute x = 0 into the equation and solve for y.
{{{y[1] = (1/2)(0)^2+2}}}
{{{y[1] = 2}}} and...
{{{y[2] = (1/2)(2)^2+2}}}
{{{y[2] = 2+2}}}
{{{y[2] = 4}}}
Now you can find the area of the first trapezoid using: {{{A = h(b[1]+b[2])/2}}}
The h here is just x = 2, so...
{{{A[1] = 2(2+4)/2}}}
{{{A[1] = 6}}}
For the second trapezoid, you have:{{{b[1] = 4}}} and {{{b[2] = 6.5}}} from:
{{{y[1] = (1/2)(2)^2+2}}}
{{{y[1] = 4}}} and...
{{{y[2] = (1/2)(3)^2+2}}}
{{{y[2] = (1/2)(9)+2}}}
{{{y[2] = 6.5}}}
So, the area of the second trapezoid is:
{{{A[2] = (1)(4+6.5)/2}}}
{{{A[2] = 10.5/2}}}
{{{A[2] = 5.25}}}
Finally, add these two areas together to get:
{{{A[t] = 6+5.25}}}
{{{A[t] = 11.25}}} as the approximate area of the enclosed region.