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Let $y=x(x-2)^2$.

Solve for $\int {x}$ $dy$ from 0 to $\frac {9}{8}$

It is known that a point, (0.5,1.125) passes through the curve.

This is the original question at part (iv)enter image description here

We have not learnt how to use substitution in integrals. This is evident usage in method 2:enter image description here. So, method 2 is not available for us. We used method 1. For method 1, this graph is used to assist to visualised thisenter image description here Where by the purple area is the area to be found and the pink area is the same as the purple area. So, I took the area of rectangle - $\int x(x-2)^2$ from 0 to $\frac {1}{2}$ which got me the answer, $\frac {41}{192}$

[MAIN FOCUS] But speaking about finding purple area, which is the answer to that question, why can't I just do this? Why is this wrong? Note that for this working, the integration is from 0 to $\frac {9}{8}$ I dont know the format for defined integral.

$\int {x}$ $dy$

$=$ $\int {y(y-2)^2}$ $dy$

$=$ $\int {\frac {y^4}{4} - \frac {4y^3}{3} + 2{y^2}}$

$=$ 1.03326416, which is not the same as $\frac {41}{192}$.

Why is this wrong? Isn't this just directly finding the purple area?

An extra note: the original way I wanted to solve this is to make $x$ subject for $y=x(x-2)^2$, however making $x$ the subject.. is extremely tedious. If it is easy, I would have just make $x$ subject and substitute to the equation and solve this immediately.

ministic2001
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1 Answers1

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The idea is integrate the function with the limits $0$ and $\frac{1}{2}$ and to subtract it from the rec-tangular with $x$-values $0$ to $\frac{1}{2}$ and $y$-values $0$ to $\frac{9}{8}$. So what you have to calculate is $$\frac{1}{2}\cdot \frac{9}{8}-\int_0^{\frac{1}{2}} x(x-2)^2 dx=\frac{41}{192}$$

The area belonging to the desired integral does not go from the $x$-axis to the graph, but from the $y$-axis to the graph.

Peter
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