Area of the given region

Question

There is a graph shown and what I did first was do the definite integral for this but I got the wrong answer when I looked in the back of my book.

The equation: y=(3-x)√x , [0,3]

Answer 1

Rebecca,

 

∫xⁿ dx = xⁿ⁺¹ / (n+1)

 

y = (3 - x)√x = 3 x^1/2 - x^3/2

 

For x^1/2 , n = 1/2 and n+1 = 3/2.

 

For x^3/2 , n = 3/2 and n+1 = 5/2.

 

The indefinite integral is 3 * x^3/2 / (3/2) - x^5/2 / (5/2) = 3 * (2/3) x^3/2 - (2/5) x^5/2.

 

Cancelling the 3s in the first term and pulling 2/5 and x^1/2 out of both terms gives (2/5) * (5 x^3/2 - x^5/2) = (2/5) x^1/2 (5x - x²).

 

Evaluating this expression at x = 3 and subtracting the expression at x = 0 gives [(2/5) * √3 * (5*3 - 3²)] - [(2/5) * √0 * (5*0 - 0²)], where the expression at x = 0 is zero. You simply have to evaluate the underelined expression.

 

Michael.

Answer 2

Three possible reasons why you have the answer:

 

1) The answer key is not always trustworthy

2) They want an estimated area under the curve using equal rectangles or trapezoids

3) You made a minor error when finding the definite integral

 

 

I will perform the definite integral and you compare the answer that I give, you have, and what the book has.

 

∫(3 - x)√x dx

 

 

Using distributive property to rewrite the integral in a way that would be easier to integrate,

 

∫(3x^1/2 - x^3/2) dx

 

 

Separate the integral into a sum of integrals by their terms.

 

3∫x^1/2 dx - ∫x^3/2 dx =

 

3(2/3)x^3/2 - (2/5)x^5/2 =

 

2x^3/2 - (2/5)x^5/2

 

 

Now we plug in the bounds.

 

[2(3)^3/2 - (2/5)(3)^5/2] - 0 = [2 * √(3²) * √(3)] - [(2/5) * √(3⁴) * √(3)]

           

                                     = 6√3 - 9(2/5)√3

 

                                     = 6√3 - (18/5)√3

 

                                     = (12/5)√3