The answer to your question is pretty simple:

Arnold Odonnell

Answered 2021-06-04
Author has **18084** answers

asked 2021-09-03

Find the exact length of the curve. \(\displaystyle{x}={\frac{{{y}^{{4}}}}{{{8}}}}+{\frac{{{1}}}{{{4}{y}^{{2}}}}},{1}\leq{y}\leq{2}\)

asked 2021-11-30

Find the exact length of the curve.

\(\displaystyle{Y}=\frac{{y}^{{4}}}{{8}}+\frac{{1}}{{4}}{y}^{{2}},\) 1 less than or equal to Y less than or equal 2.

\(\displaystyle{Y}=\frac{{y}^{{4}}}{{8}}+\frac{{1}}{{4}}{y}^{{2}},\) 1 less than or equal to Y less than or equal 2.

asked 2021-09-15

Find the exact length of the curve

\(\displaystyle{x}={\frac{{{y}^{{4}}}}{{{8}}}}+{\frac{{{1}}}{{{4}{y}^{{2}}}}}\quad{1}\leq{y}\leq{2}\)

\(\displaystyle{x}={\frac{{{y}^{{4}}}}{{{8}}}}+{\frac{{{1}}}{{{4}{y}^{{2}}}}}\quad{1}\leq{y}\leq{2}\)

asked 2021-05-18

Find the exact length of the curve

\(x=\frac{y^4}{8}+\frac{1}{4y^2}\quad1\leq y\leq2\)

\(x=\frac{y^4}{8}+\frac{1}{4y^2}\quad1\leq y\leq2\)

asked 2021-12-30

Find the exact length of the curve.

\(\displaystyle{x}={\frac{{{y}^{{4}}}}{{{8}}}}+{\frac{{{1}}}{{{4}{y}^{{2}}}}},\ {1}\le{y}\le{2}\)

\(\displaystyle{x}={\frac{{{y}^{{4}}}}{{{8}}}}+{\frac{{{1}}}{{{4}{y}^{{2}}}}},\ {1}\le{y}\le{2}\)

asked 2021-05-16

Find the exact length of the curve.

\(y=\sqrt{x-x^2}+\sin ^{-1}\sqrt{x}\)

\(y=\sqrt{x-x^2}+\sin ^{-1}\sqrt{x}\)

asked 2021-05-23

Evaluate the line integral, where C is the given curve

C \((\frac{x}{y})ds, C: x = t^3, y = t^4, 1 \leq t \leq 2\)

C \((\frac{x}{y})ds, C: x = t^3, y = t^4, 1 \leq t \leq 2\)