Integration

NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Regions above and below axis (signed areas)

Lesson

We've had a good look at ideas about approximating area, the Fundamental Theorem of Calculus and how to calculate a definite integral. Let's now take a look at how we can use these to find areas bounded by a curve and the $x$`x`-axis.

Consider the definite integral $\int_2^5\left(2x+3\right)dx$∫52(2`x`+3)`d``x`. Evaluating this integral, we get:

$\int_2^5\left(2x+3\right)dx$∫52(2x+3)dx |
$=$= | $\left[x^2+3x\right]_2^5$[x2+3x]52 |

$=$= | $\left(5^2+3\times5\right)-\left(2^2+3\times2\right)$(52+3×5)−(22+3×2) | |

$=$= | $40-10$40−10 | |

$=$= | $30$30 |

Now let's have a look at the shaded area below, which is the area between the curve $y=2x+3$`y`=2`x`+3, $x=2$`x`=2, $x=5$`x`=5 and the $x$`x`-axis:

We can find the area bound by the curve and the $x$`x`-axis using the formula for area of a trapezium:

Area | $=$= | $\frac{h\left(a+b\right)}{2}$h(a+b)2 |

$=$= | $\frac{\left(5-2\right)\left(7+13\right)}{2}$(5−2)(7+13)2 | |

$=$= | $\frac{3\times20}{2}$3×202 | |

$=$= | $30$30 |

That's exactly the same result of the definite integral $\int_2^5\left(2x+3\right)dx$∫52(2`x`+3)`d``x`.

This actually works for all curves. The definite integral of curve can be used to find the area bounded by its curve and the $x$`x`-axis.

Area bound by a curve and the x-axis

Where the curve $f\left(x\right)$`f`(`x`) lies above the $x$`x`-axis:

The area bound by the curve, the $x$`x`-axis, $x=a$`x`=`a` and $x=b$`x`=`b` is given by:

$A=\int_a^bf\left(x\right)dx$`A`=∫`b``a``f`(`x`)`d``x`

Find the area enclosed by the curve $y=-x(x-1)$`y`=−`x`(`x`−1) and the $x$`x`-axis.

We can work out the area as follows:

This result has been calculated using the definite integral, and we see we get a positive answer.

Let's see what happens if we were to calculate the area under the $x$`x`-axis.

Find the area enclosed by the curve $y=x(x-1)$`y`=`x`(`x`−1) and the $x$`x`-axis.

We can see that this graph is just a reflection of our graph from Example 1. This means our area should be exactly the same.

Let's calculate the definite integral for this one.

We can see that we actually end up with a negative value.

In fact for any region under the $x$`x`-axis, the definite integral will yield a negative value, and of course area can't be negative. So what can we do in our calculation to make sure we end up with a positive result?

We can write our definite integral with a negative out the front like this

Or we can find the absolute value of our definite integral calculation.

We can even flip the limits on our definite integral

Any of these approaches will give us the correct answer of$\frac{1}{6}$16 $units^2$`u``n``i``t``s`2

Consider the function $y=f\left(x\right)$`y`=`f`(`x`) drawn below.

Loading Graph...

Find the value of $\int_0^2f\left(x\right)dx$∫20

`f`(`x`)`d``x`.Find the value of $\int_2^5f\left(x\right)dx$∫52

`f`(`x`)`d``x`.Find the value of $\int_5^6f\left(x\right)dx$∫65

`f`(`x`)`d``x`.Hence calculate the area bounded by the function and the $x$

`x`-axis.

Consider the function $y=f\left(x\right)$`y`=`f`(`x`) drawn below.

Loading Graph...

Find the value of $\int_0^5f\left(x\right)dx$∫50

`f`(`x`)`d``x`.Find the value of $\int_5^7f\left(x\right)dx$∫75

`f`(`x`)`d``x`.Hence calculate $\int_0^7f\left(x\right)dx$∫70

`f`(`x`)`d``x`.Calculate the area bounded by the function and the $x$

`x`-axis.

Answer the following questions:

Calculate $\int_{-4}^35dx$∫3−45

`d``x`.Hence determine the area bounded by the curve $y=5$

`y`=5, the $x$`x`-axis and the bounds $x=-4$`x`=−4 and $x=3$`x`=3.

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

Apply integration methods in solving problems