For part (a), you're looking to find [tex]a[/tex] such that
[tex]\displaystyle\int_1^a\frac{\mathrm dx}{x^3}=\int_a^2\frac{\mathrm dx}{x^3}[/tex]
You have
[tex]\displaystyle\int_1^a\frac{\mathrm dx}{x^3}=-\frac1{2x^2}\bigg|_{x=1}^{x=a}=-\frac12\left(\frac1{a^2}-1\right)[/tex]
and
[tex]\displaystyle\int_a^2\frac{\mathrm dx}{x^3}=-\frac1{2x^2}\bigg|_{x=a}^{x=2}=-\frac12\left(\frac14-\frac1{a^2}\right)[/tex]
Setting these equal, you get
[tex]\displaystyle-\frac12\left(\frac1{a^2}-1\right)=-\frac12\left(\frac14-\frac1{a^2}\right)\implies a=2\sqrt{\dfrac25}[/tex]
For part (b), you have
[tex]y=\dfrac1{x^3}\implies x=\dfrac1{\sqrt[3]y}[/tex]
and you want to find [tex]b[/tex] such that
[tex]\displaystyle\int_0^{1/8}\mathrm dy+\int_{1/8}^b\frac{\mathrm dy}{\sqrt[3]y}=\int_b^1\frac{\mathrm dy}{\sqrt[3]y}[/tex]
You have
[tex]\displaystyle\int_0^{1/8}\mathrm dy+\int_{1/8}^b\frac{\mathrm dy}{y^{1/3}}=\frac18+\frac32y^{2/3}\bigg|_{y=1/8}^{y=b}=-frac14+\frac32b^{2/3}[/tex]
and
[tex]\displaystyle\int_b^1\frac{\mathrm dy}{y^{1/3}}=\frac32y^{2/3}\bigg|_{y=b}^{y=1}=\frac32-\frac32b^{2/3}[/tex]
Setting them equal gives
[tex]-\dfrac14+\dfrac32b^{2/3}=\dfrac32-\dfrac32b^{2/3}\implies b=\dfrac7{24}\sqrt{\dfrac73}\approx0.446[/tex]
