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Other AP Calculus AB unit study guides

AP Calculus AB Ultimate Guide

Unit 1: Limits and Continuity

Unit 2: Differentiation: Definition and Fundamental Properties

Unit 3: Differentiation: Composite, Implicit, and Inverse Functions

Unit 4: Contextual Applications of Differentiation

Unit 5: Analytical Applications of Differentiation

Unit 6: Integration and Accumulation of Change

Unit 7: Differential Equations

Unit 8: Applications of Integration

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Chapter 6: Definite Integrals

If *f* is continuous on the closed interval [*a*,*b*] and *F*′ = *f*, then, according to the Fundamental Theorem of Calculus

Here ∫

*baf*(*x*) is the*definite integral**of f from a to b; f*(*x*) is called the*integrand;*and*a*and*b*are called, respectively, the*lower*and*upper limits of integration*.This important theorem says that if f is the derivative of F then the definite integral of f gives the net change in F as x varies from a to b. It also says that we can evaluate any definite integral for which we can find an antiderivative of a continuous function.

By extension, a definite integral can be evaluated for any function that is bounded and piecewise continuous. Such functions are said to be integrable.

The following theorems about definite integrals are important.

**Example:**

∫2−1(3*x*2−2*x*) *dx*=*x*3−*x*2|2−1=(8−4)−(−1−1)=6

This theorem provides the tool for evaluating an infinite sum by means of a definite integral.

Suppose that a function

*f*(*x*) is continuous on the closed interval [*a*,*b*].Divide the interval into

*n*subintervals of lengths Δ*xk*(it is not necessary that the widths be of equal length, but the formulation is generally simpler if they are).Choose numbers, one in each subinterval, as follows:

*x*1 in the first,*x*2 in the second, . . .,*xk*in the*k*th, . . .,*xn*in the*n*th. Then, assuming equal width subintervals,

Δ*xk* is called a *Riemann Sum*.

Example:

Solution:

If

*f*(*x*) is nonnegative on [*a*,*b*], we see (Figure 6–1) that*f*(*xk*) Δ*x*can be regarded as the area of a typical approximating rectangle.Assuming equal width rectangles, as the number of rectangles increases, or, equivalently, as the width Δ

*x*of the rectangles approaches zero, the rectangles become an increasingly better fit to the curve.The sum of their areas gets closer and closer to the exact area under the curve.

Finally, the area bounded by the

*x*-axis, the curve, and the vertical lines*x*=*a*and*x*=*b*is given exactly by

What if *f*(*x*) is negative? Then *any area above the graph and below the x-axis is counted as negative* (Figure 6–2).

Geometrically, area is always positive, so the shaded area above the curve and below the *x*-axis equals

where the integral yields a negative number. Note that every product *f*(*xk*) Δ*x* in the shaded region is negative, since *f*(*xk*) is negative for all *x* between *a* and *b*.

It is always possible to approximate the value of a definite integral, even when an integrand cannot be expressed in terms of elementary functions.

If

*f*is nonnegative on [*a*,*b*], we interpret ∫*b a f*(*x*)*dx*as the area bounded above by*y*=*f*(*x*), below by the*x*-axis, and vertically by the lines*x*=*a*and*x*=*b*.The value of the definite integral is then approximated by dividing the area into

*n*strips, approximating the area of each strip by a rectangle or other geometric figure, then summing these approximations. We often divide the interval from*a*to*b*into*n*strips of equal width, but any strips will work.

We may approximate ∫*b a f*(*x*) *dx* by any of the following sums, where Δ*x* represents the subinterval widths:

(1)Left sum: *f*(*x*0) Δ*x*1 + *f*(*x*1) Δ*x*2 + ⋯ + *f*(*xn* − 1) Δ*xn*, using the value of *f* at the left endpoint of each subinterval.

(2)Right sum: *f*(*x*1) Δ*x*1 + *f*(*x*2) Δ*x*2 + ⋯ + *f*(*xn*) Δ*xn*, using the value of *f* at the right end of each subinterval.

(3)Midpoint sum:

using the value of *f* at the midpoint of each subinterval.

**Example :**

Approximate ∫20*x*3 *dx* by using four subintervals of equal width and calculating:

(a)the left sum

(b)the right sum

(c)the midpoint sum

(d)the integral

SOLUTIONS:

We now find the areas of the strips in Figure 6–6 by using trapezoids. We denote the bases of the trapezoids by *y*0*, y*1*, y*2*, . . ., yn* and the heights by Δ*x* = *h*1, *h*2, . . ., *hn*.

Example:

SOLUTION:

*h*=12.ℎ=12. Then,

If

*f*is an increasing function on [*a*,*b*], then*L*(*n*)⩽∫*baf*(*x*)*dx*⩽*R*(*n*), while if*f*is decreasing, then*R*(*n*)⩽∫*baf*(*x*)*dx*⩽*L*(*n*).

From Figure 6–7 we infer that the area of a trapezoid is less than the true area if the graph of *f* is concave down, but is more than the true area if the graph of *f* is concave up

Figure 6–8 is helpful in showing how the area of a midpoint rectangle compares with that of a trapezoid and with the true area.

Our graph here is concave down. If

*M*is the midpoint of*AB*, then the midpoint rectangle is*AM*1*M*2*B*. We’ve drawn*T*1*T*2 tangent to the curve at*T*(where the midpoint ordinate intersects the curve).Since the shaded triangles have equal areas, we see that area

*AM*1*M*2*B*= area*AT*1*T*2*B*. But area*AT*1*T*2*B*clearly exceeds the true area, as does the area of the midpoint rectangle. This fact justifies the right half of the inequality below; Figure 6–7 verifies the left half.

Generalizing to *n* subintervals, we conclude:

**Example:**

Write an inequality including *L*(*n*), *R*(*n*), *M*(*n*), *T*(*n*), and ∫*baf*(*t*) *dt*for the graph of *f* shown in Figure 6–9.

It is quite common to define ln *x*, the natural logarithm of *x*, as a definite integral, as follows:

This integral can be interpreted as the area bounded above by the curve *y*=1*t*(*t*>0), below by the *t*-axis, at the left by *t* = 1, and at the right by *t* = *x* (*x* > 1).

Note that if *x* = 1 the above definition yields ln 1 = 0, and if 0 < *x* < 1 we can rewrite as follows:

showing that ln *x* < 0 if 0 < *x* < 1.

With this definition of ln *x* we can approximate ln *x* using rectangles or trapezoids.

**Example :**

Show that 12<ln 2<1.12<ln 2<1.

SOLUTION:

If the function

*y*=*f*(*x*) is integrable on the interval*a*⩽*x*⩽*b*, then we define the*average value*of*f*from*a*to*b*to be

Note that (1) is equivalent to

If

*f*(*x*) ⩾ 0 for all*x*on [*a*,*b*], we can interpret (2) in terms of areas as follows:The right-hand expression represents the area under the curve of y = f(x), above the x-axis, and bounded by the vertical lines x = a and x = b.

The left-hand expression of (2) represents the area of a rectangle with the same base (b − a) and with the average value of f as its height.

**Example :**

Find the average value of *f*(*x*) = ln *x* on the interval [1,4].

SOLUTION:

If *f* is continuous on the closed interval [*a*,*b*] and *F*′ = *f*, then, according to the Fundamental Theorem of Calculus

Here ∫

*baf*(*x*) is the*definite integral**of f from a to b; f*(*x*) is called the*integrand;*and*a*and*b*are called, respectively, the*lower*and*upper limits of integration*.This important theorem says that if f is the derivative of F then the definite integral of f gives the net change in F as x varies from a to b. It also says that we can evaluate any definite integral for which we can find an antiderivative of a continuous function.

By extension, a definite integral can be evaluated for any function that is bounded and piecewise continuous. Such functions are said to be integrable.

The following theorems about definite integrals are important.

**Example:**

∫2−1(3*x*2−2*x*) *dx*=*x*3−*x*2|2−1=(8−4)−(−1−1)=6

This theorem provides the tool for evaluating an infinite sum by means of a definite integral.

Suppose that a function

*f*(*x*) is continuous on the closed interval [*a*,*b*].Divide the interval into

*n*subintervals of lengths Δ*xk*(it is not necessary that the widths be of equal length, but the formulation is generally simpler if they are).Choose numbers, one in each subinterval, as follows:

*x*1 in the first,*x*2 in the second, . . .,*xk*in the*k*th, . . .,*xn*in the*n*th. Then, assuming equal width subintervals,

Δ*xk* is called a *Riemann Sum*.

Example:

Solution:

If

*f*(*x*) is nonnegative on [*a*,*b*], we see (Figure 6–1) that*f*(*xk*) Δ*x*can be regarded as the area of a typical approximating rectangle.Assuming equal width rectangles, as the number of rectangles increases, or, equivalently, as the width Δ

*x*of the rectangles approaches zero, the rectangles become an increasingly better fit to the curve.The sum of their areas gets closer and closer to the exact area under the curve.

Finally, the area bounded by the

*x*-axis, the curve, and the vertical lines*x*=*a*and*x*=*b*is given exactly by

What if *f*(*x*) is negative? Then *any area above the graph and below the x-axis is counted as negative* (Figure 6–2).

Geometrically, area is always positive, so the shaded area above the curve and below the *x*-axis equals

where the integral yields a negative number. Note that every product *f*(*xk*) Δ*x* in the shaded region is negative, since *f*(*xk*) is negative for all *x* between *a* and *b*.

It is always possible to approximate the value of a definite integral, even when an integrand cannot be expressed in terms of elementary functions.

If

*f*is nonnegative on [*a*,*b*], we interpret ∫*b a f*(*x*)*dx*as the area bounded above by*y*=*f*(*x*), below by the*x*-axis, and vertically by the lines*x*=*a*and*x*=*b*.The value of the definite integral is then approximated by dividing the area into

*n*strips, approximating the area of each strip by a rectangle or other geometric figure, then summing these approximations. We often divide the interval from*a*to*b*into*n*strips of equal width, but any strips will work.

We may approximate ∫*b a f*(*x*) *dx* by any of the following sums, where Δ*x* represents the subinterval widths:

(1)Left sum: *f*(*x*0) Δ*x*1 + *f*(*x*1) Δ*x*2 + ⋯ + *f*(*xn* − 1) Δ*xn*, using the value of *f* at the left endpoint of each subinterval.

(2)Right sum: *f*(*x*1) Δ*x*1 + *f*(*x*2) Δ*x*2 + ⋯ + *f*(*xn*) Δ*xn*, using the value of *f* at the right end of each subinterval.

(3)Midpoint sum:

using the value of *f* at the midpoint of each subinterval.

**Example :**

Approximate ∫20*x*3 *dx* by using four subintervals of equal width and calculating:

(a)the left sum

(b)the right sum

(c)the midpoint sum

(d)the integral

SOLUTIONS:

We now find the areas of the strips in Figure 6–6 by using trapezoids. We denote the bases of the trapezoids by *y*0*, y*1*, y*2*, . . ., yn* and the heights by Δ*x* = *h*1, *h*2, . . ., *hn*.

Example:

SOLUTION:

*h*=12.ℎ=12. Then,

If

*f*is an increasing function on [*a*,*b*], then*L*(*n*)⩽∫*baf*(*x*)*dx*⩽*R*(*n*), while if*f*is decreasing, then*R*(*n*)⩽∫*baf*(*x*)*dx*⩽*L*(*n*).

From Figure 6–7 we infer that the area of a trapezoid is less than the true area if the graph of *f* is concave down, but is more than the true area if the graph of *f* is concave up

Figure 6–8 is helpful in showing how the area of a midpoint rectangle compares with that of a trapezoid and with the true area.

Our graph here is concave down. If

*M*is the midpoint of*AB*, then the midpoint rectangle is*AM*1*M*2*B*. We’ve drawn*T*1*T*2 tangent to the curve at*T*(where the midpoint ordinate intersects the curve).Since the shaded triangles have equal areas, we see that area

*AM*1*M*2*B*= area*AT*1*T*2*B*. But area*AT*1*T*2*B*clearly exceeds the true area, as does the area of the midpoint rectangle. This fact justifies the right half of the inequality below; Figure 6–7 verifies the left half.

Generalizing to *n* subintervals, we conclude:

**Example:**

Write an inequality including *L*(*n*), *R*(*n*), *M*(*n*), *T*(*n*), and ∫*baf*(*t*) *dt*for the graph of *f* shown in Figure 6–9.

It is quite common to define ln *x*, the natural logarithm of *x*, as a definite integral, as follows:

This integral can be interpreted as the area bounded above by the curve *y*=1*t*(*t*>0), below by the *t*-axis, at the left by *t* = 1, and at the right by *t* = *x* (*x* > 1).

Note that if *x* = 1 the above definition yields ln 1 = 0, and if 0 < *x* < 1 we can rewrite as follows:

showing that ln *x* < 0 if 0 < *x* < 1.

With this definition of ln *x* we can approximate ln *x* using rectangles or trapezoids.

**Example :**

Show that 12<ln 2<1.12<ln 2<1.

SOLUTION:

If the function

*y*=*f*(*x*) is integrable on the interval*a*⩽*x*⩽*b*, then we define the*average value*of*f*from*a*to*b*to be

Note that (1) is equivalent to

If

*f*(*x*) ⩾ 0 for all*x*on [*a*,*b*], we can interpret (2) in terms of areas as follows:The right-hand expression represents the area under the curve of y = f(x), above the x-axis, and bounded by the vertical lines x = a and x = b.

The left-hand expression of (2) represents the area of a rectangle with the same base (b − a) and with the average value of f as its height.

**Example :**

Find the average value of *f*(*x*) = ln *x* on the interval [1,4].

SOLUTION:

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Other AP Calculus AB unit study guides

AP Calculus AB Ultimate Guide

Unit 1: Limits and Continuity

Unit 2: Differentiation: Definition and Fundamental Properties

Unit 3: Differentiation: Composite, Implicit, and Inverse Functions

Unit 4: Contextual Applications of Differentiation

Unit 5: Analytical Applications of Differentiation

Unit 6: Integration and Accumulation of Change

Unit 7: Differential Equations

Unit 8: Applications of Integration

Studying for another AP Exam?

Check out our other AP study guides

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