Question

i want to transpose a function, \[y=243x^5\]
is this correct: \[x^5=\frac{y}{243}\Rightarrow
x=\sqrt{\frac{y^(1/5)}{3}}\]

Answer 1

\[x^5=\frac{y}{243}\Rightarrow x=\frac{\sqrt{y^{1/5}}}{3}\]

Answer 2

so that I can then integrate the area under this curve and y-axis

Answer 3

i ask, because I keep getting the wrong solution

Answer 4

\[ \int^{32}_1 \frac{y^{1/5}}{3}=[5\cdot\frac{y^{6/5}}{6\cdot3}]^{32}_1=\frac{5\cdot32\cdot2}{18}=\frac{320}{18}=17.5\]

Answer 5

However, book solution is 157.5. Can anyone help me here please?

Answer 6

Sorry, the original question and the follow both contain errors: the follow up should read
\[x^5=\frac{y}{243} \Rightarrow x=\frac{y^{1/5}}{3}\]

Answer 7

@godfreysown can u write the exact question ????????????

Answer 8

Find area between y=243x^5 and y-axis from y=1 to y=32

Answer 9

I haven't integrated correctly, that is one thing, but even if one allows for this, my solution is still out by a lot

Answer 10

I should have subtracted \[\frac{5}{18}\] from 17.5

Answer 11

but that still gets me no-where near 157.5

Answer 12

another question: how could I have rewritten the original function so as to translate the curve 45degrees to the left?

Answer 13

sorry, 90 degrees to the left

Answer 14

oh, you seem to have given up or lost connection; thanks for beginning to help anyway; cheers

Answer 15

so\[\int\limits_{1}^{32}[y^(1/5)/3]dy\]
approx :)