First slide

Evaluation of definite integrals

Question

If a function $f : [0, 27] \to R$ is differentiable then for some
$0 < α < β < 3, \int_{0}^{27} f (x) d x$ is equal to

Difficult

A

$3 [α^{2} f (α^{3}) + β^{2} f (β^{3})]$
B

$3 [α^{2} f (α) + β^{2} f (β)]$
C

$3 [α^{2} f (α^{3}) + \frac{1}{2} β^{2} f (β^{3})]$
D

$3 [α^{2} f (α) + \frac{1}{2} β^{2} f (β)]$

Solution

Let $g (x) = \int_{0}^{x^{3}} f (t) d t$ so $g (0) = 0$
$\begin{array}{l} g^{'} (x) = 3 x^{2} f (x^{3}) \\ \int_{0}^{27} f (t) d t = g (3) = \frac{g (3) - g (1)}{3 - 1} + \frac{1}{2} \frac{g (1) - g (0)}{1 - 0} \\ = g^{'} (α) + \frac{1}{2} g^{'} (β) \end{array}$
for some $α \in (1, 3), β \in (0, 1) \subseteq (0, 3),$ but $g^{'} (α) = 3 α^{2} f (α^{3})$
and $g^{'} (β) = 3 β^{2} f (β)^{3}$
Therefore, $\int_{0}^{27} f (t) d t = 3 [α^{2} f (α^{3}) + \frac{1}{2} β^{2} f (β^{3})]$

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