Chapter 4: Q25MP (page 239)

Find $\frac{d}{d x} \int_{1}^{\frac{1}{x}} \frac{e^{x t}}{t} d t$ .

Short Answer

Expert verified

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

Step by step solution

Given Information

Given that . $\frac{d}{d x} \int_{1}^{\frac{1}{x}} \frac{e^{x t}}{t} d t$ ...(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)

Solving the given equation

Compare equation (1) and equation (2).

$\frac{d}{d x} \int_{1}^{\frac{1}{x}} \frac{e^{x t}}{t} d t = \frac{e^{x (\frac{1}{x})}}{\frac{1}{x}} \times \frac{d}{d x} (\frac{1}{x}) - \frac{e^{x (1)}}{1} \times \frac{d}{d x} (1) + \int_{1}^{\frac{1}{x}} \frac{t e^{x t}}{t} d t \frac{d}{d x} \int_{1}^{\frac{1}{x}} \frac{e^{x t}}{t} d t = x e^{1} (\frac{- 1}{x^{2}}) - \frac{e^{x}}{1} (0) + \int_{1}^{\frac{1}{x}} e^{x t} d t \frac{d}{d x} \int_{1}^{\frac{1}{x}} \frac{e^{x t}}{t} d t = - \frac{1}{x} e^{1} + {[\frac{e^{x t}}{x}]}_{1}^{\frac{1}{x}}$

Therefore,

$\frac{d}{d x} \int_{1}^{\frac{1}{x}} \frac{e^{x t}}{t} d t = - \frac{1}{x} e^{1} + [\frac{e^{x (\frac{1}{x})}}{x} - \frac{e^{x (1)}}{x}] \frac{d}{d x} \int_{1}^{\frac{1}{x}} \frac{e^{x t}}{t} d t = - \frac{1}{x} e^{1} + e^{1} \frac{1}{x} - \frac{e^{x}}{x} \frac{d}{d x} \int_{1}^{\frac{1}{x}} \frac{e^{x t}}{t} d t = - \frac{e^{x}}{x}$

Hence the value of $\frac{d}{d x} \int_{1}^{\frac{1}{x}} \frac{e^{x t}}{t} d t = - \frac{e^{x}}{x}$ .

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Find $\frac{d}{d x} \int_{1}^{\frac{1}{x}} \frac{e^{x t}}{t} d t$ .