How to find integration??

Question

Given that #int_0^4 h(x) dx = 5#, evaluate:

a) #int_0^4 2h(x) dx#

b) #int_0^4 h(x)-x dx#

c) #int_0^4 h(x) +1 dx#

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Answer 1 · 2018-04-15T09:33:09

a) 10
b) -3
c) 9

a) Remember that a constant can be written outside the integral. Therefore,:
#int_0^4 2h(x) dx=2*int_0^4 h(x) dx=2*5=10#

b) Remember that #int (f(x)+g(x))=int(f(x))+int(g(x))#

Therefore:
#int_0^4 h(x)-x dx=int_0^4 h(x)dx-int_0^4x dx#
#int_0^4x dx=[1/2x^2]_0^4=16/2-1/2*0=8#
#int_0^4 h(x)dx-int_0^4x dx=5-8=-3#

c) #int_0^4 h(x) +1 dx=int_0^4 h(x)dx +int_0^4 1 dx#
#int_0^4 1 dx=[x]_0^4=4-0=4#
#int_0^4 h(x)dx +int_0^4 1 dx=5+4=9#