Given #int_0^3f(x)dx=6# and #int_0^2f(x)dx=5#. Find the value k when #int_3^2(f(x)-k)dx=5k#. How to do this question? Help needed

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Answer 1 · 2018-02-04T04:04:25

#k=-1/4#

.

#int_0^3f(x)dx=(F(x))_0^3=F(3)-F(0)=6#

#int_0^2f(x)dx=(F(x))_0^2=F(2)-F(0)=5#

If we subtract the first equation from the second one we get:

#F(2)-F(0)-F(3)+F(0)=5-6=-1#

#F(2)-F(3)=-1#

#int_3^2(f(x)-k)dx=(F(x)-kx)_3^2=F(2)-2k-(F(3)-3k)=5k#

#F(2)-F(3)-2k+3k=5k#

#-1+k=5k#

#-1=4k#

#k=-1/4#

Answer 2 · 2018-02-04T04:54:51

# -1/4#.

Prerequisites :

#(1):int_a^bphi(x)dx=int_a^cphi(x)dx+int_c^bphi(x)dx; altcltb#.

#(2):int_a^bphi(x)dx=-int_b^aphi(x)dx; altb#.

Given that, #int_0^3f(x)dx=6#,

#rArr int_0^2f(x)dx+int_2^3f(x)dx=6.........................[because, (1)]#.

#rArr 5+int_2^3f(x)dx=6..................................[because," Given]"#.

#rArr int_2^3f(x)dx=1," or, by (2), "int_3^2f(x)dx=-1.......(ast)#.

Now, #int_3^2(f(x)-k)dx=5k#,

#rArrint_3^2f(x)dx-int_3^2kdx=5k#,

#rArr -1-kint_3^2dx=5k...[because, (ast)]#,

#rArr -1-k[x]_3^2=5k#,

#rArr -1-k[2-3]=5k#,

#rArr -1-k(-1)=5k#,

#rArr -1=5k-k=4k#.

#:. k=-1/4#.