a. Explanation Given information: The curve function is y = ∫ 0 x tan t d t . The curve lies from x = 0 to x = π 6 . Calculation: Show the curve function as follows: y = ∫ 0 x tan t d t (1) Simplify Equation (1). y = ∫ 0 x sin t cos t d t (2) Let consider u = cos t . Differentiate u = cos t as follows: d u = − sin t d t Substitute u for cos t and − d u for sin t d t in the integral ∫ sin t cos t d t ∫ sin t cos t d t = − ∫ 1 u d u = − ln u Substitute cos t for u. ∫ sin t cos t d t = − ln cos t Apply the limit 0 to x as follows: ∫ 0 x tan t d t = [ − ln cos t ] 0 x = − ln cos x Differentiate Equation (1) with respect to x b. To determine Sketch the graph of the curve. c. To determine The value of the length of the curve.

Thomas' Calculus: Early Transcendentals in SI Units

14th Edition

ISBN: 9781292253114

Author: Hass, Joel R., Heil, Christopher E., WEIR, Maurice D.

Publisher: PEARSON

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In a mathematical investigation, numerical integration comprises a wide group of calculations for computing the mathematical estimation of definite integral, and likewise, the term is moreover in some cases used to depict the numerical solution of differ…

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Chapter 6.3, Problem 23E

To determine

Frame an integral for the length of the curve of the function y=∫0xtant dt.

To determine

Sketch the graph of the curve.

To determine

The value of the length of the curve.

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Chapter 6.3, Problem 22E

Chapter 6.3, Problem 24E

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Thomas' Calculus: Early Transcendentals in SI Units

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Frame an integral for the length of the curve of the function y = ∫ 0 x tan t d t .

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Frame an integral for the length of the curve of the function y=∫0xtant dt.

Sketch the graph of the curve.

The value of the length of the curve.

Want to see the full answer?

Chapter 6 Solutions