$$\text{Find area under curve} y=243x^5$$ between y=1 and y=32
So, I transpose to get integrand in y:
$$\int^{32}_1 \frac{y^{1/5}}{3}\cdot dy=17.5$$
Which may or may not be correct;
but my question is this:
How could I have rewritten the original (untransposed) f(x) so as to avoid transposition, translating the curve 90 degrees to left so as to have an integrand in x and integrate from -32 to -1 along the x-axis.
thanks for any help

Question

$$\text{Find area under curve} y=243x^5$$ between y=1 and y=32
So, I transpose to get integrand in y:
$$\int^{32}_1 \frac{y^{1/5}}{3}\cdot dy=17.5$$
Which may or may not be correct;
but my question is this:
How could I have rewritten the original (untransposed) f(x) so as to avoid transposition, translating the curve 90 degrees to left so as to have an integrand in x and integrate from -32 to -1 along the x-axis.
thanks for any help

anonymous · Answer

Thanks so kindly, but your last calculation  equates to 7/2 area units, which can't be right, intuitively, for the given area

anonymous · Answer

Sorry.

anonymous · Answer

Please see attached screenshot of geogebra file for outline of problem

anonymous · Answer

Numerically i get approximately 3.6 on excel.

anonymous · Answer

really? I'll have to learn to use excell (or my linux equivalent) didn't realize you could do such things as integration on excell; thanks