Integration (1 Viewer)

moderntortoisecat · Mar 11, 2024

answer was 4 but working out would really help

Susu123 · Mar 11, 2024

I think this is the easiest way. In the second line, I multiplied it by 2 cuz I only used the top half of the area.
I tried subtracting one line from the other and integrating in terms of the y axis, but it's too long and complicated. Hope this makes sense

Luukas.2 · Mar 11, 2024

Since the area is symmetric about the x-axis, its area can be found by doubling the integral of the function above the x-axis:

$\begin{align*} \text{The area of the property is given by } 2\int_0^1 y\,dx &= 2\int_0^1 \sqrt{9x}\,dx \qquad \text{as $y^2 = 9x \implies y = \sqrt{9x}$ above the $x$-axis} \\ &= 2 \times 3\left[\cfrac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^1 \\ &= 6 \times \frac{2}{3}\left[x\sqrt{x}\right]_0^1 \\ &= 4\left[1 - 0\right] \\ &= 4\ \text{km}^2 \qquad \text{as expected} \end{align*}$

Alternatively, integrating against the y-axis:

$\begin{align*} \text{Put $x = 1$:} \qquad y^2 &= 9x = 9 \\ y &= \pm3 \\ \\ \text{Hence, the area of the property is given by} \int_{-3}^3 1 - x\,dy &= \int_{-3}^3 1 - \frac{y^2}{9}\,dy \qquad \text{as $y^2 = 9x \implies x = \frac{y^2}{9}$} \\ &=\left[y - \frac{1}{9} \times \cfrac{y^3}{3}\right]_{-3}^3 \\ &= \left[\left(3 - \frac{3^3}{27}\right) - \left(-3 - \frac{(-3)^3}{27}\right)\right] \\ &= \left[2 - \left(-3 + \frac{27}{27}\right)\right] \\ &= 4\ \text{km}^2 \qquad \text{as expected} \end{align*}$

You could also consider that the rectangle formed by the new expressway and the y-axis is 1 km by 6 km, so 6 km² in area, and the region inside that rectangle that is not part of the property - that is, the part between the parabolic arc and the y-axis - is given by

$\begin{align*} \int_{-3}^3 x\,dy &= \int_{-3}^3 \frac{y^2}{9}\,dy \qquad \text{as $y^2 = 9x \implies x = \frac{y^2}{9}$} \\ &= \frac{1}{9}\left[\frac{y^3}{3}\right]_{-3}^3 \\ &= \frac{1}{9}\left[\frac{3^3}{3} - \frac{(-3)^3}{3}\right] \\ &= \frac{1}{9}\left(9 - \frac{-27}{3}\right) \\ &= \frac{9 + 9}{9} \\ &= 2\ \text{km}^2 \end{align*}$

Hence, the property has area of 6 km² - 2 km² = 4 km², as expected.

moderntortoisecat · Mar 15, 2024

thank you!
Another question im completely lost on seeing as theres only one Point of intersection

Average Boreduser · Mar 15, 2024

moderntortoisecat said:
thank you!
Another question im completely lost on seeing as theres only one Point of intersection
View attachment 42702

theres 2 poi- x=0, x=1

tywebb · Mar 15, 2024

moderntortoisecat said:
thank you!
Another question im completely lost on seeing as theres only one Point of intersection
View attachment 42702

I suggest you draw a diagram first.

You could do it as area between curves or just use symmetry

By symmetry,
$A=\textstyle2\int_0^1(x-x^2)\ \!dx=2[\frac{x^2}{2}-\frac{x^3}{3}]_0^1=2(\frac{1}{2}-\frac{1}{3}-0)=\frac{1}{3}$

Alternatively, by area between curves we have

$\begin{aligned}A&=\textstyle\int_0^1((x-x^2)-(x^2-x))\ \!dx\\ &=\textstyle2\int_0^1(x-x^2)\ \!dx\\ &=\textstyle2[\frac{x^2}{2}-\frac{x^3}{3}]_0^1\\ &=\textstyle2(\frac{1}{2}-\frac{1}{3}-0)\\ &=\textstyle\frac{1}{3}\end{aligned}$

moderntortoisecat · Mar 15, 2024

Oh i used the wrong equations, thanks guys!

moderntortoisecat · Mar 16, 2024

this q is from baulkham hills 2023 trials, why do we subtract the integral by 2?

I got

which i would absolute
Help

Average Boreduser · Mar 16, 2024

moderntortoisecat said:
View attachment 42719
this q is from baulkham hills 2023 trials, why do we subtract the integral by 2?

I got View attachment 42723 which i would absolute
Help

well you have to -2 because you integrated on the y-axis so there is still the square that remains in the integral that does not satisfy the exact area BOUNDED by the curve and x-axis and x=. so u minus it. also im unsure as to why you have negative in the front of the integral.

moderntortoisecat · Mar 16, 2024

wouldn't you manipulate the integral so that the derivative of the power -1 is in front of the e and then to balance it, place another (-) outside of the integral?

Average Boreduser · Mar 16, 2024

moderntortoisecat said:
wouldn't you manipulate the integral so that the derivative of the power -1 is in front of the e and then to balance it, place another (-) outside of the integral?

I started confusing myself mb. just use u-sub for this qn. adding that negative makes logic of the working look iffy.

Gantor · Mar 16, 2024

moderntortoisecat said:
View attachment 42719
this q is from baulkham hills 2023 trials, why do we subtract the integral by 2?

I got View attachment 42723 which i would absolute
Help

sub in (2-y) to the integral of 5 which can deduce the value of the shaded area minus the blank area where area is not possible to find since 1+1 is not equal 0

Luukas.2 · Mar 16, 2024

moderntortoisecat said:
View attachment 42719
this q is from baulkham hills 2023 trials, why do we subtract the integral by 2?

I got View attachment 42723 which i would absolute
Help

The problem can be approached as an area between two curves, just relative to the y-axis.

The curves that bound the region are $x=e^{2 - y}$ , which is the "top" curve, and $x = 1$ , which is the "bottom" curve. The area, against the y-axis, is then

$\begin{align*} \text{Shaded Area}\ &= \int_0^2 e^{2 - y} - 1\, dy \\ &= \left[\frac{e^{2 - y}}{-1} - y\right]_0^2 \qquad \qquad \qquad \text{noting that $\int e^{Ax + B}\, dx = \frac{1}{A}e^{Ax + B} + C$ for $A,\, B \in \mathbb{R}$ and $A\neq 0$.} \\ &= \left(\frac{e^{2 - 2}}{-1} - 2\right) - \left(\frac{e^{2 - 0}}{-1} - 0\right) \\ &= \left(-1 - 2\right) - \left(-e^2\right) \\ &= e^2 - 3\ \text{sq. units} \end{align*}$

Integration (1 Viewer)

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