Q26MP Find ddx鈭�0x2sinxttdt.... [FREE SOLUTION]

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Chapter 4: Q26MP (page 239)

Find $\frac{d}{d x} {鈭�}_{0}^{x^{2}} \frac{s i n x t}{t} d t$ .

Short Answer

Expert verified

The value of $\frac{d}{d x} {鈭�}_{0}^{x^{2}} \frac{\sin x t}{t} d t is \frac{3}{x} \sin x^{3}$ .

Step by step solution

Given Information

Given that . $\frac{d}{d x} {鈭�}_{0}^{x^{2}} \frac{\sin x t}{t} d t$ ...(1)

Formula Used

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

Solving the given equation

Compare equation (1) and equation (2).

$\frac{d}{d x} {鈭�}_{0}^{x^{2}} \frac{\sin x t}{t} d t = \frac{\sin (x . x^{2})}{x^{2}} . \frac{d}{d x} (x^{2}) - \frac{\sin (x . 0)}{0} . \frac{d}{d x} (0) + {鈭�}_{0}^{x^{2}} t \frac{\cos x t}{t} d t \frac{d}{d x} {鈭�}_{0}^{x^{2}} \frac{\sin x t}{t} d t = \frac{\sin x^{3}}{x^{2}} . (2 x) - \frac{\sin 0}{0} . (0) + {鈭�}_{0}^{x^{2}} \cos x t d t \frac{d}{d x} {鈭�}_{0}^{x^{2}} \frac{\sin x t}{t} d t = 2 x \frac{\sin x^{3}}{x^{2}} + {(\frac{\sin x t}{x})}_{0}^{x^{2}}$

Therefore,

$\frac{d}{d x} {鈭�}_{0}^{x^{2}} \frac{\sin x t}{t} d t = 2 x \frac{\sin x^{3}}{x^{2}} + \frac{\sin (x . x^{2})}{x} - \frac{\sin (x . 0)}{x} \frac{d}{d x} {鈭�}_{0}^{x^{2}} \frac{\sin x t}{t} d t = \frac{2 \sin x^{3}}{x} + \frac{\sin x^{3}}{x} \frac{d}{d x} {鈭�}_{0}^{x^{2}} \frac{\sin x t}{t} d t = \frac{2}{x} \sin x^{3} + \frac{1}{x} \sin x^{3} \frac{d}{d x} {鈭�}_{0}^{x^{2}} \frac{\sin x t}{t} d t = \frac{3}{x} \sin x^{3}$

Hence the value of $\frac{d}{d x} {鈭�}_{0}^{x^{2}} \frac{\sin x t}{t} d t = \frac{3}{x} \sin x^{3}$ .

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91影视

Find $\frac{d}{d x} {鈭�}_{0}^{x^{2}} \frac{s i n x t}{t} d t$ .

Short Answer

Step by step solution

Given Information

Formula Used

Solving the given equation

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