Question

Pls help explain this math problem?

Answer 1

`\bf \displaystyle \int\limits_(2)^(k) (6x-5)dx\implies \displaystyle \int\limits_(2)^(k)6x\cdot dx\displaystyle -\int\limits_(2)^(k)5\cdot dx\implies 6\displaystyle \int\limits_(2)^(k)x\cdot dx-\displaystyle \int\limits_(2)^(k)5\cdot dx \\\\\\ \left. 3x^2~-~5x \cfrac{}{} \right]_(2)^(k)\implies [3k^2~-~5k]~~-~~[3(2)^2~-~5(2)]~~=~~\boxed{10} \\\\\\ 3k^2-5k~~-~~[12-10]~~=~~10\implies 3k^2-5k-2=10 \\\\\\ 3k^2-5k-12=0\implies (3k+4)(k-3)=0\implies k= \begin{cases} -(4)/(3)\\\\ k=3 \end{cases}`

Answer 2

The definite integral from x=2 to x=k of (6x-5) is

x^2
6*------- - 5x, [2,k] or 3[ k^2 -5k] - 3 [ 2^2 - 5(2) ]
2 or
3[k^2 - 5k] - 3[4-10] = 10 (as given)

3k^2 -15k + 18 = 10, or
3k^2 - 15k + 8 = 0

This is a quadratic equation; solve it using the quadratic formula:
-(-15) plus or minus sqrt( [-15]^2 - 4(3)(8) )
k = ------------------------------------------------------------
6
15 plus or minus 26.36
= ------------------------------------
6