contestada

Given k > 3 and
( +
2x+2) dx = 10k
show that k3 – 10k2 + ak + b = 0
where a and bare constants to be found.

Given k gt 3 and 2x2 dx 10k show that k3 10k2 ak b 0 where a and bare constants to be found class=

Respuesta :

xKelvin

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.

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