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

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Answer 1 · 2018-04-16T06:07:06

The value of $k = 2$ $k=2$

The indefinite integral is

$\int (x + 2) d x = \frac{1}{2} x^{2} + 2 x + C$ $int(x+2)dx=1/2x^2+2x+C$

The definite integal is

$\int_{0}^{k} (x + 2) d x = {[\frac{1}{2} x^{2} + 2 x]}_{0}^{2}$ $int_0^k(x+2)dx=[1/2x^2+2x]_0^k$

$= \frac{1}{2} k^{2} + 2 k$ $=1/2k^2+2k$

But

$\int_{0}^{k} (x + 2) d x = 6$ $int_0^k(x+2)dx=6$

Therefore,

$\frac{1}{2} k^{2} + 2 k = 6$ $1/2k^2+2k=6$

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

Solving this quadratic equation in $k$ $k$

$k = \frac{- 4 \pm \sqrt{4^{2} - 4 (1) (- 12)}}{2}$ $k=(-4+-sqrt(4^2-4(1)(-12)))/(2)$

$= \frac{- 4 \pm \sqrt{64}}{2}$ $=(-4+-sqrt(64))/(2)$

$k_{1} = - 6$ $k_1=-6$

and

$k_{2} = 2$ $k_2=2$

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

Given that ∫k0(x+2)dx=6int_0^k(x+2)dx = 6 , where k>0k>0, find the value of kk.?