Chapter 1: Q12-10P (page 1)

Find $\frac{d y}{d x}$ explicitly if $y = \int_{0}^{1} \frac{e^{x u} - 1}{u} d u$ .

Short Answer

Expert verified

The value of $\frac{d y}{d x}$ is $\frac{e^{x} - 1}{x}$ .

Step by step solution

Given Information

Given that $y = \int_{0}^{1} \frac{e^{xu} - 1}{u} d u$ . ...(1)

Formula Used

We know that $\frac{d}{d x} \int_{u (x)}^{v (x)} f (x, t) d t = f (x, v) \frac{d v}{d x} - f (x, u) \frac{d u}{d x} + \int_{u}^{v} \frac{\partial f}{\partial x} d t$ ...(2)

Finding the value of dydx

Compare equation (1) and equation (2).

$\frac{d}{d x} (y) = \frac{d}{d x} \int_{0}^{1} \frac{e^{x u} - 1}{u} d u \frac{d y}{d x} = \frac{e^{x (1)} - 1}{1} \cdot \frac{d}{d x} (1) - \frac{e^{x (0)} - 1}{1} \cdot \frac{d}{d x} (0) + \int_{0}^{1} u \frac{e^{x u}}{u} d u \frac{d y}{d x} = \frac{e^{x (1)} - 1}{1} (0) - \frac{e^{x (0)} - 1}{1} (0) + \int_{0}^{1} e^{x u} d u \frac{d y}{d x} = \int_{0}^{1} e^{x u} d u$

Solving further,

$\frac{d y}{d x} = {[\frac{e^{x u}}{x}]}_{0}^{1} = \frac{1}{x} (e^{x (1)} - e^{x (0)}) = \frac{e^{x} - 1}{x}$

Therefore, $\frac{d y}{d x} = \frac{e^{x} - 1}{x}$ .

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Find $\frac{d y}{d x}$ explicitly if $y = \int_{0}^{1} \frac{e^{x u} - 1}{u} d u$ .

Short Answer

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Formula Used

Finding the value of dydx

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Step by step solution

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Most popular questions from this chapter

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Find $\frac{d y}{d x}$ explicitly if $y = \int_{0}^{1} \frac{e^{x u} - 1}{u} d u$ .