First slide

Evaluation of definite integrals

Question

If $f$ is continuously differentiable function then
$\int_{0}^{2.5} [x^{2}] f^{'} (x) d x$ is equal to

Difficult

A

$6 f (6.25) + \sum_{i \in {1, \sqrt{2}, \sqrt{3}, 2, \sqrt{5}}} f (i)$
B

$6 f (6.25) - \sum_{i \in {1, \sqrt{2}, \sqrt{3}, 2, \sqrt{5}, \sqrt{6}}}$
C

$6 f (2.5) - \sum_{i \in {1, \sqrt{2}, \sqrt{3}, 2, \sqrt{5}, \sqrt{6}}}$
D

$6 f (2.5) - \sum_{i \in {1, \sqrt{2}, \sqrt{3}, \sqrt{5}}}$

Solution

$\int_{0}^{2.5} [x^{2}] f^{'} (x) d x$
$\begin{array}{r} = \int_{0}^{1} [x^{2}] f^{'} (x) d x + \int_{1}^{\sqrt{2}} [x^{2}] f^{'} (x) + \int_{\sqrt{2}}^{\sqrt{3}} [x^{2}] f^{'} (x) d x \\ + \int_{\sqrt{3}}^{2} [x^{2}] f^{'} (x) d x + \int_{2}^{\sqrt{5}} [x^{2}] f^{'} (x) d x \\ + \int_{\sqrt{5}}^{\sqrt{6}} [x^{2}] f^{'} (x) d x + \int_{\sqrt{6}}^{25} [x^{2}] f^{'} (x) d x \\ = 0 + \int_{1}^{\sqrt{2}} f^{'} (x) d x + 2 \int_{\sqrt{2}}^{\sqrt{3}} f^{'} (x) d x + 3 \int_{\sqrt{3}}^{2} f^{'} (x) d x \\ + 4 \int_{2}^{\sqrt{5}} f^{'} (x) d x + 5 \int_{\sqrt{5}}^{\sqrt{6}} f^{'} (x) d x + 6 \int_{\sqrt{6}}^{2.5} f^{'} (x) d x \end{array}$
$\begin{array}{l} = (f (\sqrt{2}) - f (1)) + 2 (f (\sqrt{3}) - f (\sqrt{2})) \\ + 3 (f (2) - f (\sqrt{3})) + 4 (f (\sqrt{5}) - f (2)) \\ + 5 [f (\sqrt{6}) - f (\sqrt{5})] + 6 [f (2.5) - f (\sqrt{6})] \\ = 6 f (2.5) - (f (1) + f (\sqrt{2}) + f (\sqrt{3}) + f (2) \\ + (f (\sqrt{5}) + f (\sqrt{6})) \end{array}$

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