Definite integral

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\frac{d}{dx} \sin x=?\, This article/section deals with mathematical concepts appropriate for a student in late high school or early university.

A definite integral is an integral with upper and lower limits.

A definite integral.

Contents

Definite Integrals

A definite integral is the area under the curve between two points on the function. In the picture below, the yellow area is "positive" and the blue area is "negative". The integral is evaluated by adding the positive area together and subtracting the negative area.

Definiteintegralnv7.gif

If the function f(x) is real rather than complex, then the definite integral is also known as a Riemann integral.


Solving Definite Integrals

Solving a definite integral usually has two main steps: integration and subtraction.

Sometimes approximations, such as the Riemann Integral or Simpson's rule are used. These approximations are used when:

  • The exact answer is not needed, only a close approximation. (Common in Engineering)
  • The rule for integration is very complex. (Such as \int e^{x^2}\,dx)
  • The rule for integration is simply unknown. (Such as \int \zeta(x)\,dx, the Zeta function)

Example 1

Click on the image to get a better view.
A graph representation of f(x) in Example 1.
The thin blue line is f(x).
The yellow is what we add to get the value of the integral. It is the area.
The vertical green line, the lower bound, is at -3 because the lower bound of the definite integral is located at -3.
Likewise, the red line is the upper bound, which is at 5.
Click on the image to get a better view.


This is a very simple definite integral:

\int_{-3}^{5} x^2 dx

Integration


Using indefinite integration, it can be shown that:

\int x^2dx = {1 \over 3} x^3 = F(x)


Note that F(x) is the indefinite integral of f(x).

Subtraction


Now, plug 5 and − 3 into the new expression and subtract, as shown by the Fundamental Theorem of Calculus.

F(5) = {1 \over 3}5^3 = {125 \over 3}
F(-3) = {1 \over 3}(-3)^3 = {-27 \over 3}


And subtract:

{125 \over 3} - {-27 \over 3} = {125 \over 3} + {27 \over 3}
= {152 \over 3}

Example 2

This is a more complex definite integral that requires partial fractions to solve:

\int_4^{12}\frac{3x+11}{x^2-x-6}dx

Integration


See the Partial fractions in integration page for how to integrate the above expression (it is the example).
As shown on the page mentioned above:

\int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c

Subtraction

This means that we can now subtract:

\left [ 4ln|12-3|-ln|12+2|+c \right ] - \left [ 4ln|4-3|-ln|4+2|+c \right ]
= \left [ 4ln|9|-ln|14|+c \right ] - \left [ 4ln|1|-ln|6|+c \right ]
= 4ln | 9 | − ln | 14 | − 4ln | 1 | + ln | 6 |
= 4ln | 9 | − ln | 14 | + ln | 6 |
\approx 7.94160...


Note the following:

  • ln(1) = 0
  • The c on each side cancels out because we have cc
  • ln | 6 | becomes positive because it was − ( − ln | 6 | )

See Also

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