Question

Given k > 3 and

Answer 1

Answer:

See Below.

Step-by-step explanation:

We are given that:

$\displaystyle \int_3^k\left(2x+\frac{6}{x^2}\right)\, dx = 10k, k>3$

And we want to show that:

$k^3-10k^2+ak+b=0$

Integrate the definite integral:

$\displaystyle x^2-\frac{6}{x}\Big|_{3}^{k}=10k$

Evaluate:

$\displaystyle \left(k^2-\frac{6}{k}\right)-((9)-(2))\right)=10k$

Simplify:

$\displaystyle k^2-\frac{6}{k}-7=10k$

Multiply both sides by k:

$k^3-6-7k=10k^2$

Rewrite:

$k^3-10k^2-7k-6=0$

Hence, a = -7 and b = -6.