Todays Agenda Upcoming Homework Section 5.3: Evaluating Definite - - PowerPoint PPT Presentation

Today’s Agenda

• Upcoming Homework
• Section 5.3: Evaluating Definite Integrals
• Quiz

Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 25 November 2015 1 / 8

Upcoming Homework

• Written HW M: Section 5.1, #4. Section 5.2, #10,16,18,30,34,44.

Due 11/30/2015.

• WeBWorK HW 24: Section 5.3, due 12/2/2015.
• WeBWorK HW 25: Section 5.4, due 12/4/2015.
• Written HW N: Section 5.3, #54,56,60,62. Section 5.4,

#2,14,20,22,26. Due 12/4/2015.

Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 25 November 2015 2 / 8

Section 5.3

As we saw in Sections 5.1 and 5.2, evaluating definite integrals using the limit-sum definition is a long process. As it turns out, there is a much faster way to evaluate definite integrals, and the proof that it works is quite simple.

Evaluation Theorem (Fundamental Theorem of Calculus, part 1)

If f is continuous on the interval [a, b], then b

a

f (x) dx = F(b) − F(a), where F is any antiderivative of f , that is, F ′(x) = f (x).

Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 25 November 2015 3 / 8

Section 5.3

First, we claim that if a = x0 < x1 < x2 < · · · < xn−1 < xn = b is a partition of the interval [a, b], then

n

• i=1

(F(xi) − F(xi−1)) = F(b) − F(a). Observe that

n

• i=1

(F(xi) − F(xi−1)) = (F(x1) − F(x0)) + (F(x2) − F(x1)) + (F(x3) − F(x2)) + · · · + (F(xn−1) − F(xn−2)) + (F(xn) − F(xn−1)) = F(xn) − F(x0) = F(b) − F(a).

Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 25 November 2015 4 / 8

Section 5.3

Now we claim that n

i=1(F(xi) − F(xi−1)) = n i=1 f (x∗ i ) · ∆x, where x∗ i

is some number between xi−1 and xi (i.e., x∗

i ∈ [xi−1, xi]). From the Mean

Value Theorem, we know that there is some value c ∈ [xi−1, xi] such that F(xi) − F(xi−1) xi − xi−1 = F ′(c) = f (c). Letting c = x∗

i , we have that

F(xi) − F(xi−1) = (f (x∗

i ))(xi − xi−1) = f (x∗ i ) · ∆x.

Therefore,

n

• i=1

(F(xi) − F(xi−1)) =

n

• i=1

f (x∗

i ) · ∆x.

Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 25 November 2015 5 / 8

Section 5.3

Combining the results from the previous two slides, we have that F(b) − F(a) =

n

• i=1

(F(xi) − F(xi−1)) =

n

• i=1

f (x∗

i ) · ∆x.

Taking the limit of both the left and right sides of this equality, we have that lim

n→∞ F(b) − F(a) = lim n→∞ n

• i=1

f (x∗

i ) · ∆x

⇐ ⇒ F(b) − F(a) = b

a

f (x) dx, which proves the Evaluation Theorem.

Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 25 November 2015 6 / 8

Section 5.3

The Evaluation Theorem suggests a way of writing indefinite integrals, which are simply families of antiderivatives. An indefinite integral is written the same way as a definite integral, only without limits of integration:

• f (x) dx.

If F(x) is an antiderivative of f (x), then we write:

• f (x) dx = F(x) + c.

Notice that definite integrals represent an actual number that we can calculate, whereas indefinite integrals represent a family of functions that differ by only a constant.

Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 25 November 2015 7 / 8

Section 5.3

Example Problems Evaluate the following definite integrals using the Evaluation Theorem:

1 3

−2 x2 − 3 dx

2 2

0 (2x − 3)(4x2 + 1) dx

3 π

0 5ex + 3 sin x dx

4 2

1 x 2 − 2 x dx

5 1

0 x10 + 10x dx

6 √

3/2 1 √ 1−x2 dx

Find the following indefinite integrals:

7 √

x3 +

3

√ x2 dx

8

(x + 4)(2x + 1) dx

9

sin x 1−sin2 x dx

10 x2−1

x4−1 dx

Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 25 November 2015 8 / 8