First slide

Evaluation of definite integrals

Question

The difference between the greatest and he least value of the function
$F (x) = \int_{0}^{x} (t + 1) d t$ on [2, 3] is

Moderate

A

3
B

2
C

7/2
D

3/2

Solution

Differentiating the given function, we get
${F^{'} (x) = t + 1]}_{t = x} \frac{d x}{d x} - [t + 1]_{t = 0} \frac{d 0}{d x} = x + 1$
This is positive for all $x \in [2, 3],$ so $F$ is an increasing function in this interval. Therefore its greatest value is
$F (3) = \int_{0}^{3} (t + 1) d t$ and its least value is $F (2) = \int_{0}^{2} (t + 1) d t,$
so that the required difference between these values is
$\int_{0}^{3} (t + 1) d t - \int_{0}^{2} (t + 1) d t = \int_{2}^{3} (t + 1) d t = \frac{7}{2} .$

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