Let f : [0, 2] → R be a function which is continuous on [0, 2] and is differentiable on (0, 2) with f (0) = 1. Let F \left(\right. x \left.\right) = \int_{0}^{x^{2}} f \textrm{ } \left(\right. \sqrt{t} \left.\right) \textrm{ } dt for x \in [0, 2]. If F'(x) = f ' (x) for all x \in (0, 2), then F (2) equals

Engineering

Mathematics

Properties of Definite Integral

Question

Let f : [0, 2] → R be a function which is continuous on [0, 2] and is differentiable on (0, 2) with f (0) = 1. Let $F (x) = \int_{0}^{x^{2}} f (\sqrt{t}) dt$ for x $\in$ [0, 2]. If F'(x) = f ' (x) for all x $\in$ (0, 2), then F (2) equals

e – 1

e⁴ – 1

e² – 1

e⁴

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F ' (x) = f (x) · 2x = f ' (x)

$∴ \frac{f' (x)}{f (x)} = 2 x$

ln(f(x)) = x² + C

f (0) = 1 ⇒ C = 0

$f (x) = e^{x^{2}}$

Integrate $∴ F' (x) = 2 x e^{x^{2}}$

F(0) = 0 ⇒ C = – 1

$F (x) = e^{x^{2}} - 1$

F(2) = e⁴ – 1