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Introduction
1. Limits
2. Derivative basics
3. Advanced differentiation
4. Contextual uses
5. Analytical uses
6. Integration
6.1 Accumulation of change
6.2 Riemann sums & area
6.3 Definite integrals
6.4 Accumulation functions
6.5 Behavior of accumulation functions
6.6 Fundamental theorem of calculus
6.7 Indefinite integrals
6.8 u-substitution
6.9 Long division & completing the square
7. Differential equations
8. Applications of integrals
9. Testing details tag
Wrapping up
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6.6 Fundamental theorem of calculus
Achievable AP Calculus AB
6. Integration
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Fundamental theorem of calculus

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What you’ll learn:

  • Understand how definite integrals can be computed using antiderivatives
  • How to apply the Fundamental theorem of calculus part 2 to evaluate definite integrals
  • Use the reverse power rule to find basic antiderivatives

In a previous section, you saw that derivatives and integrals are opposite processes. If

F(x)=∫ax​f(t)dt,

then

F′(x)=f(x).

Because integration can “undo” differentiation, when you’re given a function f(x) you can look for an antiderivative: a function F(x) whose derivative is f(x).

That idea leads to Part 2 of the Fundamental Theorem of Calculus, which gives a practical way to compute definite integrals using antiderivatives.

Fundamental Theorem of Calculus (FTC), Part 2

If f is continuous on [a,b], and F is any antiderivative of f, then

∫ab​f(x)dx=F(x)​ab​=F(b)−F(a)

To evaluate a definite integral:

  • Find an antiderivative F of f.
  • Compute F(b)−F(a).

The vertical bar in F(x)​ab​ means “evaluate at b and subtract the value at a.” It’s a common intermediate step when working with definite integrals.

Now the key question is: how do you find antiderivatives?

Just like with derivatives, there are rules you can use. One of the most important is the reverse power rule.

Reverse power rule

Recall the power rule for derivatives:

dxd​(xn)=nxn−1

To reverse this process for integrals, add 1 to the exponent and divide by the new exponent:

∫xndx=n+1xn+1​+C

This rule requires n=−1 (otherwise you’d divide by 0). For now, we’ll avoid that case. The meaning of the +C and additional antiderivative rules are covered in the next section.

Examples

Evaluate

∫−14​(2x+3)dx

Solution

(spoiler)

Graphically, the region under f(x)=2x+3 from x=−1 to x=4 is a trapezoid. The width is 4−(−1)=5, and the parallel sides have lengths f(−1)=1 and f(4)=11. So we should expect the trapezoid’s area:

25​(1+11)=30

Now compute the same value using the FTC.

First, rewrite f(x) with explicit exponents so it’s clear how to apply the reverse power rule:

f(x)=2x1+3x0

Apply the reverse power rule term-by-term to find an antiderivative F(x):

F(x)=1+12x1+1​+0+13x0+1​+C

=x2+3x+C

Now apply the FTC:

∫−14​(2x+3)dx=(x2+3x)​−14​

=[42+3(4)]−[(−1)2+3(−1)]

=28−(−2)

=30​

Evaluate

∫02​(4x2−x​)dx

(spoiler)

Rewrite f(x) so the exponents are explicit:

f(x)=4x2−x1/2

Use the reverse power rule:

F(x)=2+14x2+1​−(1/2)+1x(1/2)+1​+C

=34​x3−32​x3/2+C

Now apply the FTC:

∫02​(4x2−x​)dx=34​x3−32​x3/2​02​

=[34​(23)−32​(23/2)]−[34​(03)−32​(03/2)]

=(332​−328​​)−(0−0)

=332−42​​​

Let f be the function whose graph is given below and let g be an antiderivative of f.

Figure 6.6.1 Graph of f
Figure 6.6.1 Graph of f

a) If g(5)=−2, find g(1).
b) Find g′(3).

Solutions

a) Find g(1) given g(5)=−2.

(spoiler)

Because g is an antiderivative of f, the FTC gives

∫15​f(x)dx=g(5)−g(1).

Solve for g(1):

g(1)=g(5)−∫15​f(x)dx

=−2−∫15​f(x)dx

Now compute the definite integral by finding the net area under the graph from x=1 to x=5. The region consists of:

  • A trapezoid on interval [1,2] with area
    • A=21​(1)(3+2)=2.5
  • A trapezoid on interval [2,5] with area
    • A=21​(2)(1+3)=4

So

∫15​f(x)dx=4+2.5=6.5

Then

g(1)=−2−6.5

=−8.5​

b) Find g′(3).

(spoiler)

Since g is an antiderivative of f, we have g′(x)=f(x).

So g′(3)=f(3). From the graph, f(3)=2​.

The table below gives values of a function g at selected points:

x g(x)
0 2
2 1
3 5
5 6

Suppose that G is an antiderivative of g such that G(0)=1. Use a right Riemann sum with the values in the table to estimate G(3).

(spoiler)

Because G is an antiderivative of g, the FTC gives

∫03​g(x)dx=G(3)−G(0).

Solve for G(3):

G(3)=G(0)+∫03​g(x)dx

We can’t compute ∫03​g(x)dx exactly from the table, so we estimate it using a right Riemann sum on the subintervals determined by the given x-values.

It may help to plot the points and draw the rectangles. Here is how it looks:

Using right endpoints to build rectangles that approximate the area under G
Using right endpoints to build rectangles that approximate the area under G

Using right endpoints:

  • On [2,3], the right endpoint is x=3, so the height is g(3)=5 and the width is 3−2=1. Area: 5⋅1=5.
  • On [0,2], the right endpoint is x=2, so the height is g(2)=1 and the width is 2−0=2. Area: 1⋅2=2.

So the right Riemann sum is 5+2=7, giving the estimate

∫03​g(x)dx≈7.

Then

G(3)=G(0)+∫03​g(x)dx

≈1+7=8​

Here’s a problem that appeared on the multiple-choice question portion of the 1998 AB exam (#11):

If f is a linear function and 0<a<b, then ∫ab​f′′(x)dx=

a) 0
b)1
c)2ab​
d)b−a
e)2b2−a2​

Solution

(spoiler)

Apply the FTC to f′′:

∫ab​f′′(x)dx=f′(b)−f′(a).

If f is linear, then

f(x)=mx+b.

Differentiate:

f′(x)=m.

So f′(a)=m and f′(b)=m, which means

f′(b)−f′(a)=m−m=0.

The correct answer choice is a) 0​.

Antiderivatives and definite integrals

  • Antiderivative F(x): satisfies F′(x)=f(x)
  • Integration “undoes” differentiation
  • Definite integral via antiderivative: ∫ab​f(x)dx=F(b)−F(a)

Fundamental theorem of calculus (FTC), part 2

  • If f is continuous on [a,b] and F is any antiderivative:
    • ∫ab​f(x)dx=F(b)−F(a)
  • To evaluate definite integrals:
    • Find antiderivative F
    • Compute F(b)−F(a)

Reverse power rule

  • For n=−1: ∫xndx=n+1xn+1​+C
  • Add 1 to exponent, divide by new exponent
  • Used term-by-term for polynomials

Worked examples

  • Apply reverse power rule to each term
  • Substitute limits into antiderivative
  • Subtract lower limit value from upper limit value

Applications with graphs and tables

  • FTC connects definite integral to net area under curve
  • Antiderivative values relate via definite integrals: g(5)−g(1)=∫15​f(x)dx
  • g′(x)=f(x) if g is antiderivative of f
  • Use Riemann sums to estimate definite integrals from tables:
    • Right Riemann sum: sum of (width × right endpoint value) for each interval

Special case: integrals of second derivatives

  • If f is linear, f′′(x)=0
  • Thus, ∫ab​f′′(x)dx=0 for any a<b

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Fundamental theorem of calculus

What you’ll learn:

  • Understand how definite integrals can be computed using antiderivatives
  • How to apply the Fundamental theorem of calculus part 2 to evaluate definite integrals
  • Use the reverse power rule to find basic antiderivatives

In a previous section, you saw that derivatives and integrals are opposite processes. If

F(x)=∫ax​f(t)dt,

then

F′(x)=f(x).

Because integration can “undo” differentiation, when you’re given a function f(x) you can look for an antiderivative: a function F(x) whose derivative is f(x).

That idea leads to Part 2 of the Fundamental Theorem of Calculus, which gives a practical way to compute definite integrals using antiderivatives.

Fundamental Theorem of Calculus (FTC), Part 2

If f is continuous on [a,b], and F is any antiderivative of f, then

∫ab​f(x)dx=F(x)​ab​=F(b)−F(a)

To evaluate a definite integral:

  • Find an antiderivative F of f.
  • Compute F(b)−F(a).

The vertical bar in F(x)​ab​ means “evaluate at b and subtract the value at a.” It’s a common intermediate step when working with definite integrals.

Now the key question is: how do you find antiderivatives?

Just like with derivatives, there are rules you can use. One of the most important is the reverse power rule.

Reverse power rule

Recall the power rule for derivatives:

dxd​(xn)=nxn−1

To reverse this process for integrals, add 1 to the exponent and divide by the new exponent:

∫xndx=n+1xn+1​+C

This rule requires n=−1 (otherwise you’d divide by 0). For now, we’ll avoid that case. The meaning of the +C and additional antiderivative rules are covered in the next section.

Examples

Evaluate

∫−14​(2x+3)dx

Solution

(spoiler)

Graphically, the region under f(x)=2x+3 from x=−1 to x=4 is a trapezoid. The width is 4−(−1)=5, and the parallel sides have lengths f(−1)=1 and f(4)=11. So we should expect the trapezoid’s area:

25​(1+11)=30

Now compute the same value using the FTC.

First, rewrite f(x) with explicit exponents so it’s clear how to apply the reverse power rule:

f(x)=2x1+3x0

Apply the reverse power rule term-by-term to find an antiderivative F(x):

F(x)=1+12x1+1​+0+13x0+1​+C

=x2+3x+C

Now apply the FTC:

∫−14​(2x+3)dx=(x2+3x)​−14​

=[42+3(4)]−[(−1)2+3(−1)]

=28−(−2)

=30​

Evaluate

∫02​(4x2−x​)dx

(spoiler)

Rewrite f(x) so the exponents are explicit:

f(x)=4x2−x1/2

Use the reverse power rule:

F(x)=2+14x2+1​−(1/2)+1x(1/2)+1​+C

=34​x3−32​x3/2+C

Now apply the FTC:

∫02​(4x2−x​)dx=34​x3−32​x3/2​02​

=[34​(23)−32​(23/2)]−[34​(03)−32​(03/2)]

=(332​−328​​)−(0−0)

=332−42​​​

Let f be the function whose graph is given below and let g be an antiderivative of f.

a) If g(5)=−2, find g(1).
b) Find g′(3).

Solutions

a) Find g(1) given g(5)=−2.

(spoiler)

Because g is an antiderivative of f, the FTC gives

∫15​f(x)dx=g(5)−g(1).

Solve for g(1):

g(1)=g(5)−∫15​f(x)dx

=−2−∫15​f(x)dx

Now compute the definite integral by finding the net area under the graph from x=1 to x=5. The region consists of:

  • A trapezoid on interval [1,2] with area
    • A=21​(1)(3+2)=2.5
  • A trapezoid on interval [2,5] with area
    • A=21​(2)(1+3)=4

So

∫15​f(x)dx=4+2.5=6.5

Then

g(1)=−2−6.5

=−8.5​

b) Find g′(3).

(spoiler)

Since g is an antiderivative of f, we have g′(x)=f(x).

So g′(3)=f(3). From the graph, f(3)=2​.

The table below gives values of a function g at selected points:

x g(x)
0 2
2 1
3 5
5 6

Suppose that G is an antiderivative of g such that G(0)=1. Use a right Riemann sum with the values in the table to estimate G(3).

(spoiler)

Because G is an antiderivative of g, the FTC gives

∫03​g(x)dx=G(3)−G(0).

Solve for G(3):

G(3)=G(0)+∫03​g(x)dx

We can’t compute ∫03​g(x)dx exactly from the table, so we estimate it using a right Riemann sum on the subintervals determined by the given x-values.

It may help to plot the points and draw the rectangles. Here is how it looks:

Using right endpoints:

  • On [2,3], the right endpoint is x=3, so the height is g(3)=5 and the width is 3−2=1. Area: 5⋅1=5.
  • On [0,2], the right endpoint is x=2, so the height is g(2)=1 and the width is 2−0=2. Area: 1⋅2=2.

So the right Riemann sum is 5+2=7, giving the estimate

∫03​g(x)dx≈7.

Then

G(3)=G(0)+∫03​g(x)dx

≈1+7=8​

Here’s a problem that appeared on the multiple-choice question portion of the 1998 AB exam (#11):

If f is a linear function and 0<a<b, then ∫ab​f′′(x)dx=

a) 0
b)1
c)2ab​
d)b−a
e)2b2−a2​

Solution

(spoiler)

Apply the FTC to f′′:

∫ab​f′′(x)dx=f′(b)−f′(a).

If f is linear, then

f(x)=mx+b.

Differentiate:

f′(x)=m.

So f′(a)=m and f′(b)=m, which means

f′(b)−f′(a)=m−m=0.

The correct answer choice is a) 0​.

Key points

Antiderivatives and definite integrals

  • Antiderivative F(x): satisfies F′(x)=f(x)
  • Integration “undoes” differentiation
  • Definite integral via antiderivative: ∫ab​f(x)dx=F(b)−F(a)

Fundamental theorem of calculus (FTC), part 2

  • If f is continuous on [a,b] and F is any antiderivative:
    • ∫ab​f(x)dx=F(b)−F(a)
  • To evaluate definite integrals:
    • Find antiderivative F
    • Compute F(b)−F(a)

Reverse power rule

  • For n=−1: ∫xndx=n+1xn+1​+C
  • Add 1 to exponent, divide by new exponent
  • Used term-by-term for polynomials

Worked examples

  • Apply reverse power rule to each term
  • Substitute limits into antiderivative
  • Subtract lower limit value from upper limit value

Applications with graphs and tables

  • FTC connects definite integral to net area under curve
  • Antiderivative values relate via definite integrals: g(5)−g(1)=∫15​f(x)dx
  • g′(x)=f(x) if g is antiderivative of f
  • Use Riemann sums to estimate definite integrals from tables:
    • Right Riemann sum: sum of (width × right endpoint value) for each interval

Special case: integrals of second derivatives

  • If f is linear, f′′(x)=0
  • Thus, ∫ab​f′′(x)dx=0 for any a<b