Find the line integral of f ( x , y ) = y e ^ x ^ 2 along the curve 𝐫 ( t ) = 4 t 𝐢 - 3

Find the line integral of f ( x , y ) = y e ^ x ^ 2 ...

Key Concepts

Line Integral

A line integral is an integral in which the function to be integrated is evaluated along a curve. It allows the computation of quantities such as work done by a force field along a path or the accumulation of a field’s intensity along a trajectory. The process involves integrating a function, often a scalar field or a vector field, over a curve in space, which is parameterized by one variable.

Curve Parameterization

Parameterizing a curve means expressing the coordinates of the points on the curve as functions of a single parameter. This allows the transformation of the line integral into an ordinary integral with respect to that parameter. The parameterization encodes both the direction and the speed at which the path is traversed.

Substitution into the Integral

Once the curve is parameterized, the function being integrated is expressed in terms of the parameter by substituting the parameterizations of the coordinates. Additionally, the differential arc length element or differential change in the parameter is computed, converting the line integral from a curve integral to a standard single-variable integral that can be evaluated using traditional integration techniques.

Transcript

00:01 Okay, let's go ahead and step through the process of how to find a line integral for the function f of x, y, equal to y times e raised to the x squared along a curve represented by r of t equal to 4ti.

00:33 Minus 3tj and t is going to go from negative 1 to 2 inclusive.

00:43 Okay, so let's don't forget the integral of the function xyds.

00:54 So that line integral is represented by the integral over the curve of the function ds with respect to s.

01:04 And d .s is given by the magnitude of this v of t times dt.

01:11 And we want to go ahead and replace the function in terms of t as well.

01:16 So we do know that v of t is given by the derivative of r.

01:23 And so we're going to take that derivative, and that will give me 4i minus 3j.

01:31 And then the magnitude of v of t is equal to the square root of four squared plus that negative three squared which is going to give me a five and so d s is equal to five d t okay and so this is going to give me the integral from negative 1 to 2 of y is negative 3t so this is going to give me a negative 3t e raised to 4 t squared so this is going to be give me a 16 t squared times 5 d t okay, so let's clean this up.

02:31 And so this is going to give me a negative 15 times the integral from negative 1 to 2 of t e raised to the 16t squared d t.

02:47 Okay, so let's go ahead and kind of work with this.

02:51 You know we're going to have to, so this was negative 15 integral from negative 1 to 2, of t, e, raised to the 16t squared d t...

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