I - The Fundamental Theorem of Calculus Lesson

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that for a continuous function f on [a, b], then equation image indicator $LaTeX: \int_a^{b}f\left(x\right)dx=F(b)-F(a)$ $\int_a^{b}f\left(x\right)dx=F(b)-F(a)$ , where F is any antiderivative of f. It is the central theorem of integral calculus and connects integration and differentiation. It enables us to compute integrals using an antiderivative of the integrand function rather than taking limits of Riemann sums. View examples of the application of the Fundamental Theorem of Calculus below.

Example 1

Find the area under the line f(x) = x + 5 from [1, 4]. text annotation indicator

Solution: $LaTeX: \int_1^{4} \left(x+5\right)dx=\frac{x^2}{2}+5x\Biggr]_1^4=\left(8+20\right)-\left(\frac{1}{2}+5\right)=22.5$ $\int_1^{4} \left(x+5\right)dx=\frac{x^2}{2}+5x\Biggr]_1^4=\left(8+20\right)-\left(\frac{1}{2}+5\right)=22.5$

solution on graph to example 1

Example 2

Evaluate equation image indicator $LaTeX: \int_1^{2} \frac{y+5y^7}{y^3}dy$ $\int_1^{2} \frac{y+5y^7}{y^3}dy$ . text annotation indicator

Solution: $LaTeX: \int_1^{2}\frac{y+5y^7}{y^3}dy=\int_1^{2}\left(\frac{1}{y^2}+5y^4\right)dy=\\ \int_1^{2}(y^{-2}+5y^4)dy=\frac{y^{-1}}{-1}+5\frac{y^5}{5}\Biggr]_1^2=-\frac{1}{2}+2^5=31.5$ $\int_1^{2}\frac{y+5y^7}{y^3}dy=\int_1^{2}\left(\frac{1}{y^2}+5y^4\right)dy=\\ \int_1^{2}(y^{-2}+5y^4)dy=\frac{y^{-1}}{-1}+5\frac{y^5}{5}\Biggr]_1^2=-\frac{1}{2}+2^5=31.5$

Example 3

Consider equation image indicator $LaTeX: \int_{-2}^{2} \left(x^4+1\right)dx$ $\int_{-2}^{2} \left(x^4+1\right)dx$ .

x to 4 plus one with shaded area on graph text annotation indicator

Solution: Notice that f(x) is symmetric with respect to the y=axis and it is an even function, i.e., f(-x)=f(x).

When f is an even function, then:

$LaTeX: \int_{-a}^{a}f\left(x\right)dx=2\int_0^{a}f(x)dx\\ \int_{-2}^{2}(x^4+1)dx=2\int_0^{2}(x^4+1)dx=\\ 2\left(\frac{x^5}{x}+5\right)\Biggr|_0^2=2\left(\frac{32}{5}+2\right)=\frac{84}{5}=16.8$ $\int_{-a}^{a}f\left(x\right)dx=2\int_0^{a}f(x)dx\\ \int_{-2}^{2}(x^4+1)dx=2\int_0^{2}(x^4+1)dx=\\ 2\left(\frac{x^5}{x}+5\right)\Biggr|_0^2=2\left(\frac{32}{5}+2\right)=\frac{84}{5}=16.8$

Example 4

Consider

equation image indicator $LaTeX: \int_{-1}^{1} \left(x^3-x\right)dx\\ \int_{-1}^{1} \left(x^3-x\right)dx=\frac{x^4}{4}=\frac{x^2}{2}\biggr|_{-1}^1 =\left(\frac{1}{4}-\frac{1}{2}\right)-\left(\frac{\left(-1\right)^4}{4}-\frac{\left(-1\right)^2}{2}\right)=0$ $\int_{-1}^{1} \left(x^3-x\right)dx\\ \int_{-1}^{1} \left(x^3-x\right)dx=\frac{x^4}{4}=\frac{x^2}{2}\biggr|_{-1}^1 =\left(\frac{1}{4}-\frac{1}{2}\right)-\left(\frac{\left(-1\right)^4}{4}-\frac{\left(-1\right)^2}{2}\right)=0$

graph with shaded area text annotation indicator

Solution: Notice that it is an odd function, i.e., f(-x)=-f(x). When f is an odd function then: $LaTeX: \int_{-a}^{a}f\left(x\right)dx=0$ $\int_{-a}^{a}f\left(x\right)dx=0$

View the presentation below illustrating how the Fundamental Theorem of Calculus is applied to finding the area under a curve.

View the presentation below from the beginning to 6:35.

Mean Value Theorem for Integrals

Recall that the area of a region under a curve is greater than the area of an inscribed rectangle and less than the area of a circumscribed rectangle. It follows that somewhere between the inscribed and circumscribed rectangles, there is a rectangle whose area is precisely equal to the area under a curve. It is this idea that forms the basis of the Mean Value Theorem for Integrals:

If f is continuous on [a, b], then there exists a number c in [a, b] such that equation image indicator $LaTeX: \int_a^{b}f\left(x\right)dx=f(c)(b-a)$ $\int_a^{b}f\left(x\right)dx=f(c)(b-a)$ .

Geometrically, for positive functions f, there is a number c such that the rectangle with base [a, b] and height f(c) has the same area as the region under the graph of f from a to b. Note that the Mean Value Theorem for Integrals is a consequence of the Mean Value Theorem for derivatives and the Fundamental Theorem of Calculus.

Average Value of a Function

Finding the average value of a set of n numbers is identical to finding the average value of a set of n functional values

f(c₁), f(c₂),..., f(c_n-1), f(c_n), equation image indicator $LaTeX: \frac{ \sum_{i=1}^nf(c_i)}{n}$ $\frac{ \sum_{i=1}^nf(c_i)}{n}$ . If you partition [a, b] into n subintervals of equal width $LaTeX: \Delta x=\frac{b-a}{n}$ $\Delta x=\frac{b-a}{n}$ and then multiply and divide the average value of the set of n functional values by (b - a), the result is

equation image indicator $LaTeX: \frac{ \sum_{i=1}^nf(c_i)}{n}=\frac{ \sum_{i=1}^nf(c_i)}{n}=\frac{1}{b-a}\sum_{i=1}^nf(c_i)\frac{b-a}{n}=\frac{1}{b-a}\cdot \sum_{i=1}^nf(c_i)\Delta x$ $\frac{ \sum_{i=1}^nf(c_i)}{n}=\frac{ \sum_{i=1}^nf(c_i)}{n}=\frac{1}{b-a}\sum_{i=1}^nf(c_i)\frac{b-a}{n}=\frac{1}{b-a}\cdot \sum_{i=1}^nf(c_i)\Delta x$

Taking the limit as equation image indicator $LaTeX: n \to \infty$ $n \to \infty$ produces .

Thus the average value of an integrable function f on the closed interval [a, b] is equation image indicator $LaTeX: f_{avg}=\frac{1}{b-a} \int_a^{b}f\left(x\right)dx$ $f_{avg}=\frac{1}{b-a} \int_a^{b}f\left(x\right)dx$ .

Notice that the average value of f on [a, b] is the integral of f divided by the length of the interval. If f is continuous and nonnegative on [a, b], its average value suggests an average height of the graph of f. Interpreting an integral as the area of a rectangle under the curve's graph, then the rectangle's height times its width is its area. Thus, equation image indicator $LaTeX: f_{avg}\cdot\left(b-1\right)= \int_a^{b}f\left(x\right)dx$ $f_{avg}\cdot\left(b-1\right)= \int_a^{b}f\left(x\right)dx$ .

Links to an external site.Explore and practice the average value of a function concept by CLICKING HERE. Links to an external site.

Accumulated Change from a Rate of Change

The accumulated change in a quantity whose rate of change is measured by the function f(x) over an interval a ≤ x ≤ b is equal to the area under the graph of the function f(x) over the interval [a, b]. To conceive of an accumulation function defined in x is to imagine a total accumulated area for each value of x. In general, any time a Riemann sum is calculated for a rate of change, an approximation of the total change results. It may be helpful to think of a Riemann sum as the accrual of incremental portions, each of which is composed multiplicatively of two quantities f(c) and Δx.

5 accumulation pics on graphs

Recall that velocity is really the rate at which the position of an object is changing. Then the accumulated change is the total change in position, which is how far the object moved. Click HERE to view the accumulated change involving velocity and position presentation Links to an external site..

Rate of change functions are meant to be continuous, and accumulated change computations for such functions are found by computing an area under the corresponding curve. Functions that record discrete counts of a quantity, however, are only approximately modeled by continuous functions. The accumulated values of count data are found not by computing area under the curve, but simply by adding the appropriate counts.

Links to an external site. Links to an external site.Explore the relationship of the accumulation function (antiderivative) to the area under the curve by CLICKING HERE. Links to an external site.

Net Change Theorem

The integral of a function's rate of change over an interval a < x < b, without regard to possible sign change of the function on [a, b] is the net change. Such a rate of change could be either positive, negative, or a combination of both. The Net Change Theorem expresses this relationship as equation image indicator $LaTeX: \int_a^{b}F'\left(x\right)dx=F(b)-F(a)$ $\int_a^{b}F'\left(x\right)dx=F(b)-F(a)$ , where F'(x) = f(x).

Thinking in terms of area under a curve, if an integrable function f is non-positive, the terms equation image indicator $LaTeX: f\left(c_k\right)\Delta_k$ $f\left(c_k\right)\Delta_k$ in the Riemann sums for f(x) = F'(x) over an interval [a, b] are all negatives of rectangle areas. The limit of Riemann sums, which is the integral of f from a to b, is the negative of the area of the region between the graph of f and the x-axis, or

equation image indicator $LaTeX: \int_a^{b}f\left(x\right)dx$ $\int_a^{b}f\left(x\right)dx$ = - (area under the curve) when f(x) < 0. Another way of writing this is Area = $LaTeX: - \int_a^{b}f\left(x\right)dx$ $- \int_a^{b}f\left(x\right)dx$ when f(x) < 0. In general, for any integrable function, equation image indicator $LaTeX: \int_a^{b}f\left(x\right)dx$ $\int_a^{b}f\left(x\right)dx$ = (area above the x-axis) - (area below the x-axis). View an example of computing net area text annotation indicator and total area.

Example 1: Net Area and Total Area

Determine the total and net area of the region bounded by equation image indicator $LaTeX: f\left(x\right)=4-x^2$ $f\left(x\right)=4-x^2$ , and the x- and y-axes, and the line x = 3.

net area versus total area depicted on graph

Net Area:

$LaTeX: \int_a^{b}f\left(x\right)dx= \int_0^{3} \left(4-x^2\right)dx=4x-\frac{x^3}{3} \Bigg]_0^3= 12-9=3$ $\int_a^{b}f\left(x\right)dx= \int_0^{3} \left(4-x^2\right)dx=4x-\frac{x^3}{3} \Bigg]_0^3= 12-9=3$

Total Area:

$LaTeX: \int_a^{b}f\left(x\right)dx=\left|A_1\right|+\left|A_2\right|+...\\ A_1=\int_0^{2}(4-x^2)dx=4x-\frac{x^3}{3}\Bigg]_2^0=8-\frac{8}{3}=\frac{16}{3}\\ A_2=\int_0^{2}(4-x^2)dx=4x-\frac{x^3}{3}\Bigg]_2^3\\ (12-9)-(8-\frac{8}{3})=-\frac{7}{3}\\ A_Total=|A_1|+|A_2|+\Big|\frac{16}{3}\Big|+\Big|-\frac{7}{3}\Big|=\frac{23}{3}$ $\int_a^{b}f\left(x\right)dx=\left|A_1\right|+\left|A_2\right|+...\\ A_1=\int_0^{2}(4-x^2)dx=4x-\frac{x^3}{3}\Bigg]_2^0=8-\frac{8}{3}=\frac{16}{3}\\ A_2=\int_0^{2}(4-x^2)dx=4x-\frac{x^3}{3}\Bigg]_2^3\\ (12-9)-(8-\frac{8}{3})=-\frac{7}{3}\\ A_Total=|A_1|+|A_2|+\Big|\frac{16}{3}\Big|+\Big|-\frac{7}{3}\Big|=\frac{23}{3}$

Example 2: Net and Total Change

A particle moves along a line so that its velocity at time t is equation image indicator $LaTeX: v\left(t\right)=t^2-t-2$ $v\left(t\right)=t^2-t-2$ (measured in $LaTeX: \frac{m}{\sec}$ $\frac{m}{\sec}$ ). Find the displacement (net change) of the particle during the time period 1 < t < 4. Find the distance traveled (total distance) during this time period.

net total change with velocity depicted on graph text annotation indicator

Example 3

net area versus total area

net area versus total area: the net area is equal to the value of the integral connections

area (otherwise noted as total area): this equals the "actual" area. You have to find the total of all areas above the x-axis, then find the absolute value of the areas below the x-axis and add those together

Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus states that if f(x) is continuous on an open interval I containing a, then for every x in the interval, equation image indicator $LaTeX: \frac{d}{dx}\left[\int_a^{x}f\left(t\right)dt\right]=f\left(x\right)$ $\frac{d}{dx}\left[\int_a^{x}f\left(t\right)dt\right]=f\left(x\right)$ .

Explore the application of the Second Fundamental Theorem of Calculus by CLICKING HERE Links to an external site..

Fundamental Theorem of Calculus Practice

Use antiderivatives to compute the definite integral.

Fundamental Theorem of Calculus: Even More Problems!

Complete problems from your textbook and/or online resources as needed to ensure your complete understanding of increasing and decreasing functions.

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