Answer:
See Below.
Step-by-step explanation:
We are given that:
[tex]\displaystyle \int_3^k\left(2x+\frac{6}{x^2}\right)\, dx = 10k, k>3[/tex]
And we want to show that:
[tex]k^3-10k^2+ak+b=0[/tex]
Integrate the definite integral:
[tex]\displaystyle x^2-\frac{6}{x}\Big|_{3}^{k}=10k[/tex]
Evaluate:
[tex]\displaystyle \left(k^2-\frac{6}{k}\right)-((9)-(2))\right)=10k[/tex]
Simplify:
[tex]\displaystyle k^2-\frac{6}{k}-7=10k[/tex]
Multiply both sides by k:
[tex]k^3-6-7k=10k^2[/tex]
Rewrite:
[tex]k^3-10k^2-7k-6=0[/tex]
Hence, a = -7 and b = -6.
