Kenneth S. answered • 09/22/16
Tutor
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Calculus will seem easy if you have the right tutor!
When sketched, the line has y-intercept 3 & x-intercept 6.
And the other curve begins at the origin and curves upward & to the right, with downward concavity.
The intersection of these two curves occurs when x = 1. To represent a finite enclosed area, I'll use the y-axis as the final boundary.
Integrate from 0 to 1 on (-x/2 + 3 -2.5√x)dx and you will get, I hope, 3 - ¼ - 5/3) which is 7/12.
Check this carefully...a tutor is possibly just as liable to a silly unobserved mistake as anyone.
Kenneth S.
Of course it's wrong...I see that I neglected to use the upper boundary y=3.
Your remark that there is no answer is wrong, too.
The problem must be done by expressing the integrand as 3 - (expression for the line) and integrating from 0 to 1. Then there's another part, that must have the integrand 3 - 2.5 sqrtx), integrated from 1 until the point where the two curves intersect, which you find by setting the integrand to zero, i.e. 3 = 2.5sqrt(x).
I believe this sets you onto the correct path.
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09/23/16
Gregory R.
09/22/16