# The Definite Integral

## Key Questions

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#### Explanation:

Above the x-axis
If $f \left(x\right)$ is non-negative over the interval $\left[a , b\right]$, then
${\int}_{a}^{b} f \left(x\right) \mathrm{dx}$ is the area of the region between the graph and the $x$-axis between $x = a$ and $x = b$

Below the x-axis
If $f \left(x\right)$ is non-positive over the interval $\left[a , b\right]$, then
${\int}_{a}^{b} f \left(x\right) \mathrm{dx}$ is the negative of area of the region between the graph and the $x$-axis between $x = a$ and $x = b$

Example:

${\int}_{0}^{3} \sqrt{9 - {x}^{2}} \mathrm{dx}$.

$a = 0$ and $b = 3$.
$f \left(x\right) = \sqrt{9 - {x}^{2}}$ is never negative, so it is not negative on $\left[0 , 3\right]$

Therefore the integral is equal to the area under the curve and above the $x$-axis between $x = 0$ and $x = 3$.

The graph of $y = \sqrt{9 - {x}^{2}}$ is the part of ${y}^{2} = 9 - {x}^{2}$ that has non-negative $y$-values. It is the upper semicircle for ${x}^{2} + {y}^{2} = 9$
The part between $x = 0$ and $x = 3$ is a quarter of a circle with radius 3.

graph{y = sqrt(9-x^2)*(sqrt(1.5^2-(x-1.5)^2))/(sqrt(1.5^2-(x-1.5)^2)) [-2.63, 6.137, -0.812, 3.572]}

So
${\int}_{0}^{3} \sqrt{9 - {x}^{2}} \mathrm{dx}$ is $\frac{1}{4}$ of the area of the circle with radius $3$
${\int}_{0}^{3} \sqrt{9 - {x}^{2}} \mathrm{dx} = \frac{9 \pi}{4}$

An indefinite integral of a function $f \left(x\right)$ is a family of functions $g \left(x\right)$ for which: $g ' \left(x\right) = f \left(x\right)$

#### Explanation:

An indefinite integral of a function $f \left(x\right)$ is a family of functions $g \left(x\right)$ for which: $g ' \left(x\right) = f \left(x\right)$.

Examples:

1) if $f \left(x\right) = {x}^{3}$, then indefinite integral is:

$\int {x}^{3} \mathrm{dx} = {x}^{4} / 4 + C$, because:

$\left({x}^{4} / 4\right) ' = 4 \cdot {x}^{3} / 4 = {x}^{3}$, and $C ' = 0$ for any real constant $C$

2) If $f \left(x\right) = \cos x$, then

$\int \cos x \mathrm{dx} = \sin x + C$, because:

$\left(\sin x\right) ' = \cos x$

• A definite integral is when you integrate a function over a specified interval. When completed you have a definite answer.

Definite Integral because it is bounded

${\int}_{0}^{1} 3 x \mathrm{dx}$ evaluates to $\left[\frac{3 {\left(1\right)}^{2}}{2} - \frac{3 {\left(0\right)}^{2}}{2}\right] = \frac{3}{2} - 0 = \frac{3}{2} = 1.5$

Indefinite Integral because it is not bounded

$\int 3 x \mathrm{dx}$ evaluates to $\frac{3 {x}^{2}}{2} + C$