Consider the function f(x)=(x)/(x^2+1). (a) Find the signed area between the graph of f(x) and t

Consider the function f(x)=(x)/(x^2+1). (a) Find the signed...

Key Concepts

Integration Techniques

Integration techniques, such as substitution, are used to evaluate integrals that involve complex functions. These methods simplify the integration process by transforming the integrand into a more manageable form, which is essential when working with rational functions or other non-standard integrals.

Intersection Points

Intersection points of functions are the x-values where the graphs meet. They are important when determining the correct limits of integration for computing areas between curves, as these points often delineate regions where one function is above or below the other.

Area Between Curves

To find the area between two curves, one integrates the absolute difference between the functions over a given interval. This technique computes the total area enclosed between the curves without cancellation from areas above and below one another.

Signed Area

The signed area between a graph and the x-axis is determined by integrating the function over a specific interval, where areas above the axis contribute positively and areas below contribute negatively. This concept is crucial for interpreting the net effect of the function over an interval.

Definite Integral

A definite integral calculates the net accumulation of a function’s values over an interval and is used to find areas under a curve. In this context, it computes the signed area between a function and the x?axis between given limits.

Transcript

00:01 Of the region between the graph of f of x and the x -axis between x is negative 2 and x is negative 1.

00:09 So here is the graph of x cubed plus 1.

00:13 Here is x is negative 2, here is x equals negative 1.

00:18 So here's the x -axis.

00:20 Notice the function, the graph is below the x -axis, so the function is negative on our interval.

00:26 So we want to find this area between the graph of the function, the x -axis, between x is negative 2 and negative 1.

00:36 So we're looking for this area in here.

00:38 Normally the area will be equal to the definite integral, but because our function is negative on this interval, the area is going to equal the negative of the definite integral.

00:50 So area is going to equal the negative of the integral from negative to negative to negative to negative 1 ,000, of x cubed plus 1 d x.

01:05 So to evaluate this definite integral, we need the antiderivative of x cubed plus 1.

01:10 That will be x to the 4 over 4 plus x.

01:16 And then we need to evaluate this between negative 2 and negative 1.

01:29 So this expression, when x is negative 1, we're going to have 1.

01:36 Fourth plus x which is plus negative one or minus one.

01:41 Then we need to subtract this expression evaluated when x is negative two.

01:47 Negative two to the fourth will be 16 divided by four is four plus x plus negative two.

01:54 Same thing is minus two...

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