Given that #int_1^3f(x)dx=3# and #int_3^6f(x)dx=5#, find: #int_1^3[f(x)+x]dx# ?

Question

and in my book the solution is
#int_1^3[f(x)+x]dx#
#=int_1^3f(x)dx+int_1^3xf(x)#
#=3+[x^2/2]^3 =7#
What is the confusion?

but when im solving it im getting 4 as an answer

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Answer 1 · 2018-02-03T20:45:18

See below.

We have:

#=int_1^3 f(x) dx+ int_1^3 x dx#

We know from the question that #int_1^3 f(x) dx = 3#. Also, from the basic power rule of integration, we have #int_a^b x dx = [1/2x^2]_1^3#.

#=3 + [1/2x^2]_1^3#

#=3 + 1/2(3)^2 - 1/2(1)^2#

#=3 + 9/2 - 1/2#

#=3 + 8/2#

#=3 + 4#

#=7#

Hopefully this helps!