Q14P Given that 鈭�0鈭�-dxy2+x2=蟺2y,... [FREE SOLUTION]

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Chapter 4: Q14P (page 237)

Given that ${鈭�}_{0}^{鈭�} - \frac{d x}{y^{2} + x^{2}} = \frac{蟺}{2 y}$ , differentiate with respect to y and so evaluate ${鈭�}_{0}^{鈭�} - \frac{dx}{y^{2} + x^{2}} .$

Short Answer

Expert verified

The value of ${鈭�}_{0}^{鈭�} - \frac{dx}{y^{2} + x^{2}}$ is $\frac{蟺}{4 y^{3}}$ .

Step by step solution

Given Information

Given that ${鈭�}_{0}^{鈭�} - \frac{dx}{y^{2} + x^{2}} = \frac{蟺}{2 y}$ .

Formula Used

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

Evaluate ∫0∞-dx(y2+x2).

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

But,

${鈭�}_{0}^{鈭�} \frac{d x}{y^{2} + x^{2}} d x = \frac{蟺}{2 y} {鈭�}_{0}^{鈭�} \frac{d x}{y^{2} + x^{2}} d x = \frac{d}{d y} (\frac{蟺}{2 y}) = - \frac{蟺}{2 y^{2}}$

Therefore,

$\frac{- 蟺}{2 y^{2}} - = - 2 y {鈭�}_{0}^{鈭�} \frac{d x}{{(y^{2} + x^{2})}^{2}} {鈭�}_{0}^{鈭�} \frac{d x}{{(y^{2} + x^{2})}^{2}} = \frac{- 蟺}{(2 y^{2}) (- 2 y^{2})} {鈭�}_{0}^{鈭�} \frac{d x}{{(y^{2} + x^{2})}^{2}} = \frac{蟺}{4 y^{3}}$

Hence ${鈭�}_{0}^{鈭�} \frac{d x}{{(y^{2} + x^{2})}^{2}} = \frac{蟺}{4 y^{3}}$ .

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91影视

Given that ${鈭�}_{0}^{鈭�} - \frac{d x}{y^{2} + x^{2}} = \frac{蟺}{2 y}$ , differentiate with respect to y and so evaluate ${鈭�}_{0}^{鈭�} - \frac{dx}{y^{2} + x^{2}} .$

Short Answer

Step by step solution

Given Information

Formula Used

Evaluate ∫0∞-dx(y2+x2).

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91影视

Given that 鈭�0鈭�-dxy2+x2=蟺2y, differentiate with respect to y and so evaluate鈭�0鈭�-dxy2+x2.

Short Answer

Step by step solution

Given Information

Formula Used

Evaluate ∫0∞-dx(y2+x2).

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Particle Model of Matter

Electricity

Electrostatics

Electricity and Magnetism

Famous Physicists

Magnetism and Electromagnetic Induction

Study anywhere. Anytime. Across all devices.

Given that ${鈭�}_{0}^{鈭�} - \frac{d x}{y^{2} + x^{2}} = \frac{蟺}{2 y}$ , differentiate with respect to y and so evaluate ${鈭�}_{0}^{鈭�} - \frac{dx}{y^{2} + x^{2}} .$