The bounds m and M used in the Bounds on an Integral Theorem for $\int _ { - 1 } ^ { 2 } \left[ 2 - ( x - 1 ) ^ { 2 } \right] d x$ are

A) $- 1,2$
B) $- 2,1$
C) $- 2 , - 1$
D) $- 2,2$
E) $- 2,3$

Question

The bounds m and M used in the Bounds on an Integral Theorem for $\int _ { - 1 } ^ { 2 } \left[ 2 - ( x - 1 ) ^ { 2 } \right] d x$ are

A) $- 1,2$
B) $- 2,1$
C) $- 2 , - 1$
D) $- 2,2$
E) $- 2,3$

Quizplus · Accepted Answer

The bounds m and M in the Bounds on an Integral Theorem refer to the minimum and maximum values of the function on the given interval, respectively. For the function $f(x) = 2 - (x - 1)^2$, we need to find its minimum and maximum values on the interval $[-1, 2]$. The function is a downward-opening parabola with its vertex at $x = 1$, which lies within the interval. At $x = 1$, $f(x) = 2 - (1 - 1)^2 = 2$, which is the maximum value (M). The minimum value (m) occurs at the endpoints of the interval, which are $x = -1$ and $x = 2$. At $x = -1$, $f(x) = 2 - ((-1) - 1)^2 = 2 - 4 = -2$, and at $x = 2$, $f(x) = 2 - (2 - 1)^2 = 2 - 1 = 1$. Therefore, the minimum value on the interval is $-2$, and the maximum value is $2$, making the correct bounds $-2, 2$.

The Bounds M and M Used in the Bounds on an Integral