Find the interval [a,b] for which the value of the integral ...

Question

∫_a ^b (2+x-x²)dx is a maximum. Please explain your reasoning.

Answer 1

The graph of y = f(x) =  -x²+x+2 = (-x + 2 )(x + 1) is a parabola opening downward.  The graph has x-intercepts at x = -1 and x = 2. Between x = -1 and x = 2, the graph lies above the x-axis.  Otherwise, the graph lies below the x-axis.

 

 Let R be the region bounded by the graph of y = f(x) and the x-axis between x = a and x = b.  The definite integral from x=a to x=b is the area of the part of R that lies above the x-axis minus the area of the part of R that lies below the x-axis.  

 

So, the definite integral of f(x) from x = a to x = b will be maximized when a = -1 and b = 2.  

    

Answer 2

Set the derivative of the integral equal to zero.

 

2 + x - x² = 0

 

0 = x² - x - 2

 

 

Solve for x.

 

0 = (x - 2)(x + 1)

 

x = -1         and        x = 2

 

These are the values of the critical points.

 

 

Set the second derivative of the integral equal to zero.

 

1 - 2x = 0

 

x = 1/2

 

This is the inflection point.  If we evaluate the second derivative before and after this point, we can see where the graph of the integral is concave down.  That will give us the range of interval where the maximum lies.

 

1 - 2(0) = 1 - 0 = 1   --->  concave up

1 - 2(1) = 1 - 2 = -1  --->  concave down

 

Knowing this, the integral has a maximum in the interval (1/2, 2).