The graph of f consists of line segments, as shown in the figure. Evaluate each definite integra

The graph of f consists of line segments, as shown in the...

Key Concepts

Definite Integral as Signed Area

The definite integral is understood as the net signed area between a function and the x-axis over a certain interval. Areas above the x-axis contribute positively to the integral, while those below contribute negatively. This interpretation allows one to evaluate integrals by computing the areas of simple geometric shapes when the function has a piecewise or easily defined structure.

Geometric Evaluation of Integrals

When a function is composed of line segments or simple curves, its graph can often be broken into basic geometric shapes such as triangles, rectangles, or trapezoids. By applying standard area formulas to these shapes, the definite integral can be calculated directly without resorting to antiderivatives.

Piecewise Linear Functions

A piecewise linear function is one defined by different linear expressions over various subintervals of its domain. Since the graph is constructed from straight-line segments, its area under the curve over any interval can be computed by breaking the region into simple, well-known geometric figures.

Constant Multiple Rule in Integration

The constant multiple rule states that multiplying a function by a constant scales the definite integral by that same constant. This property simplifies the integration of functions that are scaled by a constant since one can first compute the integral of the original function and then multiply the result by the constant.

Additivity of Integrals

The additivity property of the definite integral allows one to split the integral over a larger interval into a sum of integrals over smaller, non-overlapping subintervals. This is particularly useful for piecewise functions where different rules apply on different intervals, enabling the overall integral to be computed by summing the areas from each segment.

Transcript

00:01 We are given graphs of functions consisting of line segments, and we're asked to evaluate each definite integral by using geometric formulas.

00:13 So we're given the function f consisting of line segments, and in part a, we're asked to find the integral from 0 to 1 of negative f of x, dx.

00:38 Well, to understand the geometry, we see that negative f of x, well, this is going to be the area of a triangle.

01:10 In particular, it's the area of a triangle with a base of one and a height of one.

01:16 So this is one -half times base 1 times the height 1, which is 1 -half.

01:25 The reason the negative is included is because the function is below the x -axis.

01:30 Then in part b, we're asked to find the integral from 3 to 4 of 3f of x dx.

01:44 Well, we can use homogeneity to pull the scalar out of the integral.

01:50 So this is three times, and then we have the integral from three to four of f of x.

01:55 But this is the area, again, of another triangle.

02:00 In particular, from three to four, i'm sorry, not a triangle.

02:04 This is the area of a rectangle with horizontal length of one and vertical length of two.

02:17 So this is three times the horizontal or the vertical two times the horizontal one for a total of six.

02:32 Then in part c we're asked to find the integral from zero to seven of f of x dx.

02:49 Here we're going to use the addativity of the integral.

02:53 This is the same as the integral from zero to one of f of x plus the integral from one from one.

03:16 1 to 6 of f of x d x plus the integral from 6 to 7 of f of x d x.

03:27 Now notice that the first and the last integrals, these are the opposites since the function is below the x -axis, of the area of a triangle.

03:38 So this is going to be the opposite of the area of a triangle times 2.

03:44 In particular, both these triangles have bases and heights of length 1.

03:53 And then the integral from 1 to 6, notice that the function in the x -axis forms a trapezoid.

04:01 So this is the area of a trapezoid, and the trapezoid has base lengths of 1 and 6 minus 1 or 5.

04:15 And so this is equal to negative 2 times the area of the triangle, which is 1⁄2 times base 1 times height 1, plus the area of the trapezoid, which is one half times the sum of the bases, which was, as i said before, 1 and 6 minus 1 or 5, times the height of the trapezoid, which is 2.

04:49 And so this simplifies to negative 1 plus 6 or positive 5.

05:06 Then in part b, we're asked to find the integral from 5 to 11 of f of x dx.

05:26 Once again, we can find this integral by breaking it up into smaller, simpler integrals.

05:31 This is the same as the integral from 5 to 6 of f of x plus the integral from 6 to 10 of f of x, plus the integral from 10 to 11 of fx dx.

05:52 Now notice that the first and third integrals, these are the areas of two triangles, each with a base and height of one...

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