Given that $int_0^k(x+2)dx = 6$ , where $k>0$ , find the value of $k$ .?

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Answer 1 · 2018-04-16T06:07:06

The value of $k=2$

The indefinite integral is

$int(x+2)dx=1/2x^2+2x+C$

The definite integal is

$int_0^k(x+2)dx=[1/2x^2+2x]_0^k$

$=1/2k^2+2k$

But

$int_0^k(x+2)dx=6$

Therefore,

$1/2k^2+2k=6$

$k^2+4k-12=0$

Solving this quadratic equation in $k$

$k=(-4+-sqrt(4^2-4(1)(-12)))/(2)$

$=(-4+-sqrt(64))/(2)$

$k_1=-6$

and

$k_2=2$

As $k>0$ , $k=2$

Given that int_0^k(x+2)dx = 6int_0^k(x+2)dx = 6 , where k>0k>0, find the value of kk.?