This question aims to find the length of the curve by applying** line integral** along the curve.

It is difficult to find the exact equation of the function along the **curve** so we need a certain formula to find the exact measurements. **Line integral** solves this problem as it is a type of integration that is performed on the functions present **along the curve.**

The line integral along the curve is also called **path integral** or **curve integral**. It can be found by finding the **sum** of all the points present on the curve with some **differential vector** along the curve.

The values of x and y are given and these are:

\[x = e^t + e^{- t}\]

\[y = 5 – 2t \]

The limits are as follows:

\[0 \leq t \leq 4 \]

## Expert Answer

By using the formula to find the length $ l $ of the curve:

\[L = \int_{a}^{b} \sqrt { (\frac { dx } { dt } ) ^ 2 + (\frac { dy } { dt } ) ^ 2 } \, dt \]

\[\frac{dx}{dt} = e^t – e^{-t}\]

\[\frac{dy}{dt} = -2\]

\[L = \int_{0}^{4} \sqrt { ( e ^ t – e ^ {-t} ) ^ 2 + ( – 2 ) ^ 2 } \, dt \]

\[L = \int_{0}^{4} \sqrt { e ^ 2t – 2 + e ^ {-2t} + 4 } \, dt \]

\[L = \int_{0}^{4} \sqrt { ( e ^ t – e ^ {-t} ) ^ 2 } \, dt \]

\[L = \int_{0}^{4} \sqrt { e ^ t – e ^ {-t} } \, dt \]

\[L = [ e ^ t – e ^ { -t } ] ^ { 4 } _ {0} dt \]

\[L = e ^ 4 – e ^ { -4 } – e ^ 0 + e ^ 0 \]

\[L = e ^ 4 – e ^ { -4 }\]

## Numerical Results

**The length $ L $ of the curve is $ e ^ 4 – e ^ { -4 } $.**

**Ex**ample

Find the length of the curve if the limits are $ \[0 \leq t \leq 2\].

\[L = \int_{a}^{b} \sqrt {(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt \]

\[\frac{dx}{dt} = e^t – e^{-t}\]

\[\frac{dy}{dt} =- 2\]

\[L = \int_{0}^{2} \sqrt { ( e ^ t – e ^ {-t} )^2 + (-2)^2}\, dt\]

\[L = \int_{0}^{2} \sqrt {e^2t – 2 + e^{-2t} + 4 }\, dt\]

\[L = \int_{0}^{2} \sqrt {(e^t – e^{-t} )^2 }\, dt\]

\[ L = \int_{0}^{2} \sqrt { e ^ t – e ^ {-t} } \, dt \]

By putting the limits:

\[ L = e ^ 2 – e ^ { -2 } – e ^ 0 + e ^ 0 \]

\[ L = e ^ 2 – e ^ { -2 }\]

**The length $ L $ of the curve is $ e ^ 2 – e ^ { -2} $**

*Image/Mathematical drawings are created in Geogebra.*