Definite Integrals, The Fundamental Theorem of Calculus Parts 1 and 2 And the Mean Value Theorem for Integrals

When we find the area under a curve by adding rectangles, the answer is called a Riemann sum. subinterval partition The width of a rectangle is called a subinterval. The entire interval is called the partition. Subintervals do not all have to be the same size. v t

subinterval partition If the partition is denoted by P, then the length of the longest subinterval is called the norm of P and is denoted by. Our book calls this max As gets smaller, the approximation for the area gets better. Therefore, the exact area is: if P is a partition of the interval v t

is called the definite integral of over. If we use subintervals of equal length, then the length of a subinterval is: The definite integral is then given by:

Leibnitz introduced a simpler notation for the definite integral: Note that the very small change in x becomes dx.

Integration Symbol lower limit of integration upper limit of integration integrand variable of integration (dummy variable) It is called a dummy variable because the answer does not depend on the variable chosen.

We have the notation for integration, but we still need to learn how to evaluate the integral.

time velocity After 4 seconds, the object has gone 12 feet. Consider an object moving at a constant rate of 3 ft/sec. Since rate. time = distance: If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.

If the velocity varies: Distance: ( C=0 since s=0 at t=0 ) After 4 seconds: The distance is still equal to the area under the curve! Notice that the area is a trapezoid. Note that s is the antiderivative Of the velocity, v.

What if: We could split the area under the curve into a lot of thin trapezoids, and each trapezoid would behave like the large one in the previous example. It seems reasonable that the distance will equal the area under the curve which will be the antiderivative of the velocity function.

The area under the curve Why can we use anti-derivatives to find the area under a curve?

Let’s see: Let area under the curve from a to x. (“ a ” is a constant) Then:

min f max f Consider the area under the curve on the previous slide from x to x+h. The area of a rectangle drawn under the curve would be less than the actual area under the curve. The area of a rectangle drawn above the curve would be more than the actual area under the curve. h

As h gets smaller, min f and max f get closer together and approach f(x). That means that : This is the definition of derivative! Take the anti-derivative of both sides to find an explicit formula for area. initial value -- the area from a to a = 0 Where F(x) is an an antiderivative of f(x)

As h gets smaller, min f and max f get closer together. Area under curve of y= f(x) from a to x is equal to the antiderivative at x minus antiderivative at a. Where F(x) is an an antiderivative of f(x) Therefore :

Area Now we have three expressions to represent the area under the curve or the Definite Integral.

Area from x=0 to x=1 Example: Find the area under the curve from x = 1 to x = 2. Area from x=0 to x=2 Area under the curve from x = 1 to x = 2. Let’s Practice using the new technique The vertical line means to evaluate the function at the upper limit and then Subtract the value of the function at the lower limit

Example: Find the area under the curve from x = 1 to x = 2. To do the same problem on the TI-83 or TI-84:

The Fundamental Theorem of Calculus Part I If a function is continuous on the closed interval [a, b], Then the definite integral where F is any function such that F ’ (x) = f(x) for any x in The interval [a, b]. Let’s review and extend :

Examples 1 – 0 = 1 Keep in mind that if the function being integrated is positive, we are finding the area between the curve and the x axis.

EXAMPLE Evaluate the definite integral.

SOLUTION

EXAMPLE Evaluate.

SOLUTION The definite integral can be used to find the area between the graph and the x-axis. But area is usually considered to be a nonnegative quantity. However, when the graph is below the x-axis, the computed value of the definite integral will be negative. Let’s see what we need to do to find the area between a curve and the x axis. Why do we get a negative?

Example – finding the area between curve and x-axis: Find the area between the x-axis and the curve from to. On the TI-83 or TI-84: Note: If you use the absolute value function, you don’t need to find the roots (x-intercepts). pos. neg.

Integration with Absolute Value inside the integral. We need to rewrite the integral into two parts using the definition of absolute value. 2

Ex. Find the area of the region bounded by y = 2x 2 – 3x + 2, the x-axis, x = 0, and x = 2 (no problem here because function is always positive on [0,2]). Just do the definite integral.

...Rules for Definite Integrals

Examples

Example

The Fundamental Theorem of Calculus Part II If f is continuous on an open interval “I” containing a, then for every x in the interval “I” Examples. Apply the Second Fundamental Theorem of Calculus When you differentiate a function that you just integrated from a constant to x, you get the original function back. Note that the x is not on top. What can we do?

The Mean Value Theorem for Integrals If f is continuous on [a, b], then there exists a number c in the open interval (a, b) such that: inscribed rectangle Mean Value rect. Mean Value rectangle area is equal to actual area under curve. Circumscribed Rect a b a c b a b Area under Curve = height X width

Find the value c guaranteed by the Mean Value Theorem for Integrals for the function f(x) = x 3 over [0, 2]. 4 = 2c 3 c 3 = c