State Standard – 16.0a Students use definite integrals in problems involving area. Objective – To be able to use the 2 nd derivative test to find concavity and points of inflection.

For Instance: We can easily calculate the exact area of regions with straight sides. l w A = lw b h A = ½bh However, it isn’t so easy to find the area of a region with curved sides.

Inscribed rectangles are all below the curve: Lower Sum Circumscribed rectangles are all above the curve: Upper sum

Lower Sum Upper Sum

Let’s find the area under the curve by using rectangles using right endpoints. –3–2– y = x 2 from [0,4] What is the Area of each rectangle? A = (1)(1 2 )+(1)(2 2 )+(1)(3 2 )+(1)(4 2 ) A = A = 30 sq units n = 4

Let’s find the area under the curve by using rectangles using left endpoints. –3–2– y = x 2 from [0,4] What is the Area of each rectangle? A = (1)(0 2 )+(1)(1 2 )+(1)(2 2 )+(1)(3 2 ) A = A = 14 sq units n = 4

–3–2– A = 30 sq units –3–2– A = 14 sq units A = ( ) / 2 A = 22 sq units

Let’s find the area under the curve by using rectangles using left endpoints. –3–2– y = x 2 from [0,4] What is the Area of each rectangle? A =+++ = 17.5 sq units n =

Approximate the area from (a) the left side (b) the right side 1) 2) 3)

State Standard – 13.0a Students know the definition of the definite integral using by using Riemann sums. Objective – To be able to use definite integrals to find the area under a curve.

Let’s find the area under the curve by using rectangles using right endpoints. –3–2– y = x 2 from [0,4] What is the Area of each rectangle? A = (1)(4 2 )+(1)(3 2 )+(1)(2 2 )+(1)(1 2 ) A = A = 30 sq units n = 4 We could also use a Right-hand Rectangular Approximation Method (RRAM).

Let’s find the area under the curve by using rectangles using left endpoints. –3–2– y = x 2 from [0,4] What is the Area of each rectangle? A = (1)(0 2 )+(1)(1 2 )+(1)(2 2 )+(1)(3 2 ) A = A = 14 sq units n = 4 This is called the Left-hand Rectangular Approximation Method (LRAM).

–3–2– y = x 2 from [0,4] What is the Area of each rectangle? A =(1)(0.5 2 ) +(1)(1.5 2 ) +(1)(2.5 2 ) +(1)(3.5 2 ) A = A = 21 sq units n = 4 Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM).

Mid-Point Rule: ( h = width of subinterval )

Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM). Approximate area: In this example there are four subintervals. As the number of subintervals increases, so does the accuracy.

Approximate area: width of subinterval With 8 subintervals:

Approximate the area from the midpoint MRAM 1) 2) 3)

State Standard – 13.0a Students know the definition of the definite integral using by using Riemann sums. Objective – To be able to use definite integrals to find the area under a curve.

–3–2– –3–2– –3–2– RRAM LRAM MRAM Right Rectangular Approximation Method

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. Definite Integrals Definition Area Under a Curve (as a Definite Integral) If y = f(x) is nonnegative and can be integrated over a closed interval [a,b], then the area under the curve y = f(x) from a to b is the integral of f from a to b,

If the velocity varies: Distance: After 4 seconds: The distance is still equal to the area under the curve! Notice that the area is a trapezoid.

1. If the upper and lower limits are equal, then the integral is zero. 2. Reversing the limits changes the sign. 3. Constant multiples can be moved outside. Integrals can be added and subtracted. Properties of Definite Integrals

5. Integrals can be separated

Given

p. 391 # 21 – 25 all

5.3 The Fundamental Theorem of Calculus Morro Rock, California

The First Fundamental Theorem of Calculus If f is continuous at every point of, and if F is any antiderivative of f on, then (Also called the Integral Evaluation Theorem)

Integrals such as are called indefinite integrals because we can not find a definite value for the answer. When finding indefinite integrals, we always include the “plus C”.

Integrals such as are called definite integrals because we can find a definite value for the answer. The constant always cancels when finding a definite integral, so we leave it out!

Area from x=0 to x=1 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.

Actual area under curve:

Find the area between the x-axis and the curve from to. pos. neg.

5.3 Homework Pg – 30 and 35

5.3b The 2 nd Fundamental Theorem of Calculus Morro Rock, California

The 2 nd Fundamental Theorem of Calculus If f is continuous on, then the function has a derivative at every point in, and

Second Fundamental Theorem: 1. Derivative of an integral.

2. Derivative matches upper limit of integration. Second Fundamental Theorem: 1. Derivative of an integral.

2. Derivative matches upper limit of integration. 3. Lower limit of integration is a constant. Second Fundamental Theorem:

1. Derivative of an integral. 2. Derivative matches upper limit of integration. 3. Lower limit of integration is a constant. New variable. Second Fundamental Theorem:

1. Derivative of an integral. 2. Derivative matches upper limit of integration. 3. Lower limit of integration is a constant. The long way: Second Fundamental Theorem:

1. Derivative of an integral. 2. Derivative matches upper limit of integration. 3. Lower limit of integration is a constant.

The upper limit of integration does not match the derivative, but we could use the chain rule.

The lower limit of integration is not a constant, but the upper limit is. We can change the sign of the integral and reverse the limits.

5.3b Homework Pg. 402 # 7 – 17

5.5 Integration by Substitution

The chain rule allows us to differentiate a wide variety of functions, but we are able to find antiderivatives for only a limited range of functions. We can sometimes use substitution to rewrite functions in a form that we can integrate.

Example 1: The variable of integration must match the variable in the expression. Don’t forget to substitute the value for u back into the problem!

Example 2: One of the clues that we look for is if we can find a function and its derivative in the integrand. The derivative of is. Note that this only worked because of the 2x in the original. Many integrals can not be done by substitution.

Example 3: Solve for dx.

Example 4:

Example 6: We solve for because we can find it in the integrand. 2 x dx

5.5 Homework Pg. 420 – – 5, 7 – 14, 19, 21, 23, 25, and 31