Question

Given k > 3 and

( +
2x+2) dx = 10k
show that k3 – 10k2 + ak + b = 0
where a and bare constants to be found.

Answer 1

See Below.

Explanation:

We are given that:

`\displaystyle \int_3^k\left(2x+(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-(6)/(x)\Big|_(3)^(k)=10k`

Evaluate:

`\displaystyle \left(k^2-(6)/(k)\right)-((9)-(2))\right)=10k`

Simplify:

`\displaystyle k^2-(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.