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Fundamental Theorem of Calculus

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Fundamental Theorem of Calculus
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Velocity due to gravity can be easily calculated by the formula: v = gt , where g is the acceleration due to gravity (9.8m/s 2 ) and t is time in seconds. In fact, a decent approximation can be calculated in your head easily by rounding 9.8 to 10 so you can just add a decimal place to the time.

Using this function for velocity, how could you find a function that represented the position of the object after a given time? What about a function that represented the instantaneous acceleration of the object at a given time?

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- KhanAcademy - The Indefinite Integral

Guidance

If you think that evaluating areas under curves is a tedious process you are probably right. Fortunately, there is an easier method. In this section, we shall give a general method of evaluating definite integrals (area under the curve) by using antiderivatives.

Definition: The Antiderivative
If F ' ( x ) = f ( x ), then F '( x ) is said to be the antiderivative of f ( x ).

There are rules for finding the antiderivatives of simple power functions such as f ( x ) = x 2 . As you read through them, try to think about why they make sense, keeping in mind that differentiation reverses integration.

Rules of Finding the Antiderivatives of Power Functions
  • The Power Rule
\int x^n dx = \frac{1} {n + 1} x^{n + 1} + C
where C is constant of integration and n is a rational number not equal to -1.
  • A Constant Multiple of a Function Rule
\int k x^n dx = k \int x^n dx = k \cdot \frac{1} {n + 1} x^{n + 1} + C
where k is a constant.
  • Sum and Difference Rule
\int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx
  • The Constant rule
\int k \cdot dx = kx + C
where k is a constant. (Notice that this rule comes as a result of the power rule above.)

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus makes the relationship between derivatives and integrals clear. Integration performed on a function can be reversed by differentiation.

The Fundamental Theorem of Calculus
If a function f ( x ) is defined over the interval [ a , b ] and if F ( x ) is the antidervative of f on [ a, b ], then
\int_{a}^{b} f(x) dx = F(x)|^b_a
= F(b) - F(a)

We can use the relationship between differentiation and integration outlined in the Fundamental Theorem of Calculus to compute definite integrals more quickly.

Example A

Evaluate \int_{1}^{2} x^2 dx.

Solution

This integral tells us to evaluate the area under the curve f ( x ) = x 2 , which is a parabola over the interval [1, 2], as shown in the figure below.

To compute the integral according to the Fundamental Theorem of Calculus, we need to find the antiderivative of f ( x ) = x 2 . It turns out to be F ( x ) = (1/3) x 3 + C , where C is a constant of integration. How can we get this? Think about the functions that will have derivatives of x 2 . Take the derivative of F ( x ) to check that we have found such a function. (For more specific rules, see the box after this example). Substituting into the fundamental theorem,

\int_{a}^{b} f(x) dx = F(x)|^b_a
\int_{1}^{2} x^2 dx = \left [\frac{1} {3} x^3 + C\right ]^2_1
= \left [\frac{1} {3} (2)^3 + C\right ] - \left [\frac{1} {3} (1)^3 + C\right ]
= \left [\frac{8} {3} + C\right ] - \left [\frac{1} {3} + C\right ]
= \frac{7} {3} + C - C
= \frac{7} {3}

So the area under the curve is (7/3) units 2 .

Example B

Evaluate \int x^3dx.

Solution

Since \int x^n dx = \frac{1} {n + 1}x^{n + 1} + C , we have

\int x^3 dx = \frac{1} {3 + 1}x^{3 + 1} + C
= \frac{1} {4}x^4 + C
To check our answer we can take the derivative of  \frac{1} {4}x^4 + C and verify that it is \,\! x^3 , the original function in our integral.

Example C

Evaluate \int 5x^2 dx.

Solution

Using the constant multiple of a power rule, the coefficient 5 can be removed outside the integral:

\int 5x^2 dx = 5 \int x^2 dx
Then we can integrate:
= 5 \cdot \frac{1} {2 + 1} x^{2 + 1} + C
= \frac{5} {3} x^3 + C
Again, if we wanted to check our work we could take the derivative of \frac{5} {3} x^3 + C and verify that we get \,\! 5x^2

Vocabulary

The Fundamental Theorem of Calculus demonstrates that integration performed on a function can be reversed by differentiation.

Integrals allow for the calculation of the area between a line (such as the x-axis) and a curve, or of the area between two curves. Since the area is generally given in square units, it is technically only an approximation, but can be an effectively infinitely close one!

The antiderivative has much the same relationship to a function that a square root has to a constant. The antiderivative of a function is the function whose derivative is the function you want the antiderivative of.

Guided Practice

Questions

1) Evaluate \int (3x^3 - 4x^2 + 2)dx.

2) Evaluate \int_{2}^{5} \sqrt{x} dx.

3)Use the fundamental theorem of calculus to solve: \int_{4}^{6} \frac{dx}{x}

4)Use the fundamental theorem of calculus to solve: \int_{-2p}^{2p} 3cos(x) dx =

5)Use the 2nd fundamental theorem of calculus to solve: A(x) = \int_{3}^{x} cot^3 (t) dt

Solutions

1) Using the sum and difference rule we can separate our integral into three integrals:

\int (3x^3 - 4x^2 + 2)dx =
3 \left( \int x^3 dx \right) - 4 \left( \int x^2 dx \right) + \left( \int 2dx \right)

\to 3 \cdot \frac{1} {4} x^4 - 4 \cdot \frac{1} {3}x^3 + 2x + C \to \frac{3} {4} x^4 - \frac{4} {3} x^3 + 2x + C

2) The evaluation of this integral represents calculating the area under the curve y = \sqrt{x} from x = -2 to x = 3, shown in the figure below.

\int_{2}^{5} \sqrt{x} dx = \int_{2}^{5} x^{1/2} dx
= \left [\frac{1} {\frac{1} {2} + 1} x^{1/2 + 1}\right ]^5_2
= \left [\frac{1} {3/2} x^{3/2}\right ]^5_2
= \frac{2} {3} \left [x^{3/2}\right ]^5_2
= \frac{2} {3}\left [5^{3/2} - 2^{3/2}\right ]
= 5.57
So the area under the curve is 5.57.

3) Given what we know, that if F(x) = ln x, then F'(x) = \frac {1}{x}

Thus, we apply the fundamental theorem of calculus:
\int-{4}^{6} \frac{dx}{x} = ln x |-{4}{6}
= F(6) - F(4) = [ln(6)] - [ln(4)] = 0.4055

4) Given what we know, that if F(x) = 3sin(x), then F'(x) = 3cos(x)

So we apply the fundamental theorem of calculus:
\int_{-2p}^{2p} 3cos dx = 3sin(x) | _{-2p}^{2p}
= F(8) - F(0) = [3sin(2p)] - [3sin(-2p)] = 1 - 0 = 0

5) Find A'(\frac{3p}{4})

The second theorem states: \int_{3}^{x}cot^3 (t) dt is the anti-derivative of cot^3(x)
So, A'(x) = cot^3(x)
Substituting in x = \frac{3p}{4} we get an answer of A'{4} = -1

Practice

Evaluate the Integral:

  1. Evaluate the integral \int_{0}^{3} 5xdx
  2. Evaluate the integral \int_{0}^{1} x^4dx
  3. Evaluate the integral \int_{1}^{4} (x - 3)dx

Find the Integral:

  1. Find the integral of ( x + 1)(2 x - 3) from -1 to 2.
  2. Find the integral of \sqrt{x} from 0 to 9.
  3. Find the integral of \int_{-1}^{0} - 3 dx =
  4. Find the integral of \int_{-1}^{3} dx =
  5. Find the integral of \int_{-p}^{\frac{p}{2}} - 4cos(x) dx =
  6. Find the integral of \int_{0}^{2} -dx =
  7. Find the integral of \int_{2}^{7} \frac{dx}{x}
  8. Find the integral of \int_{-2}^{0}x + 5 dx =
  9. Find the integral of \int_{-p}^{\frac{3p}{2}}6sin(x) dx =
  10. Find the integral of \int_{6}^{7} \frac{dx}{x}

Challenge yourself:

  1. Sketch y = x 3 and y = x on the same coordinate system and then find the area of the region enclosed between them (a) in the first quadrant and (b) in the first and third quadrants.
  2. Evaluate the integral \int_{-R}^{R} (\pi R^2 - \pi x^2) dx where R is a constant.

Apply the second theorem of calculus:

  1. A(x) = \int_{2}^{x} tan^3(t) dt
  2. Find: \frac{d}{dx} \int_{1}^{x} csc^2(t) dt
  3. Find: \frac{d}{dx} \int_{-2}^{x^2} sec(t) dt

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