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#