I require some help integrating this function for a modelling project and not really sure where to start. $$f(x)=(-0.53\sqrt[3]{x-16.7}+2.8)^2$$ I started by converting the decimals into fractions to see if it would be easier, however I am stuck. Fraction form: $$f(x)=(-\frac {53\sqrt[3]{x-\frac{167}{10}}}{100}+\frac{14}{5})^2$$
I have also tried to take out the denominators in order to simplify the question however I am not sure I have done that correctly. I believe I can take out the $5$ from $\frac{14}{5}$ as $\frac{1}{25}$ and the 100 under the cube root as $10^4$. I also think I can take out the $10$ from inside the cube root to give me $10^{2\over3}$. This means the function can be rewritten as: $${1\over25*10^{14\over3}}(- {265\sqrt[3]{10x-{167}}}+14*10^{7\over3})^2 $$ Would this be right and where would I go from here? Can I also take out negative one to give me: $${{1*-1^2}\over25*10^{14\over3}}({265\sqrt[3]{10x-{167}}}-14*10^{7\over3})^2 $$
Differentiation and integration are linear. Unlike performing arithmetic, writing the constants in terms of decimals or fraction won't change your calculation.
The constants are really distractions. Write
$$ f(x)=(a\sqrt[3]{x-b}+c)^2. $$ Let $u=a\sqrt[3]{x-b}+c$. Then $$ (\frac{u-c}{a})^3+b=x,\quad dx = 3(\frac{u-c}{a})^2\frac{1}{a}\;du $$
Now you can calculate $\int f(x)\;dx$ by a u-substitution: $$ \int f(x)\;dx = \int u^2\cdot3(\frac{u-c}{a})^2\frac{1}{a}\;du =\frac{3}{a^3}\int u^2(u-c)^2\;du $$ which is easy since you now have a polynomial to integrate.