Definite integral

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

A definite integral.

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.

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.

Example 1

This is a very simple definite integral:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{-3}^{5} x^2 dx}

Integration

Using indefinite integration, it can be shown that:

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

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:

$\left[4ln|12-3|-ln|12+2|+c\right]-\left[4ln|4-3|-ln|4+2|+c\right]$

$=\left[4ln|9|-ln|14|+c\right]-\left[4ln|1|-ln|6|+c\right]$

$=4ln|9|-ln|14|-4ln|1|+ln|6|$

$=4ln|9|-ln|14|+ln|6|$

$\approx 6.02941...$

Note the following:

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

See Also

• Integral
• Indefinite integral