If f is continuously differentiable function then∫01.5 x2f′(x)dx is

∫01.5 x2f′(x)dx=∫01 x2f′(x)dx+∫12 x2f′(x)dx+∫21.5 x2f′(x)dx=0+∫12 f′(x)dx+2∫21.5 f′(x)dx=f(2)−f(1)+2[f(1.5)−f(2)]=2f(1.5)−f(2)−f(1)

If f is continuously differentiable function then∫01.5 x2f′(x)dx is

∫01.5 x2f′(x)dx=∫01 x2f′(x)dx+∫12 x2f′(x)dx+∫21.5 x2f′(x)dx=0+∫12 f′(x)dx+2∫21.5 f′(x)dx=f(2)−f(1)+2[f(1.5)−f(2)]=2f(1.5)−f(2)−f(1)

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If $f$ is continuously differentiable function then
$\int_{0}^{1.5} [x^{2}] f^{'} (x) d x$ is

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$2 f (1.5) + f (\sqrt{2}) + f (1)$

$f (1.5) + f (\sqrt{2}) + f (1)$

$2 f (1.5) - f (\sqrt{2}) - f (1)$

$f (1.5) - f (\sqrt{2}) - f (1)$

answer is D.

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Detailed Solution

$\begin{array}{l} \int_{0}^{1.5} [x^{2}] f^{'} (x) d x \\ = \int_{0}^{1} [x^{2}] f^{'} (x) d x + \int_{1}^{\sqrt{2}} [x^{2}] f^{'} (x) d x + \int_{\sqrt{2}}^{1.5} [x^{2}] f^{'} (x) d x \\ = 0 + \int_{1}^{\sqrt{2}} f^{'} (x) d x + 2 \int_{\sqrt{2}}^{1.5} f^{'} (x) d x \\ = f (\sqrt{2}) - f (1) + 2 [f (1.5) - f (\sqrt{2})] \\ = 2 f (1.5) - f (\sqrt{2}) - f (1) \end{array}$