Given that #int_0^3f(x)dx=6# and #int_0^2f(x)dx=5#. Show that #int_2^3f(x)dx#. How to do this?

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1 Answer
Alvin L.
Feb 3, 2018

#int_2^3 f(x)\ dx=1#

Explanation:

If we consider an integral as the area under a curve, we can say that the area under the curve on the interval #[2,3]# can be expressed as the area under the curve on the interval #[0,3]# minus the area under the curve on the interval #[0,2]#.

You can think of it as taking the area under #[0,3]# and then removing the area #[0,2]#, which would leave you with the area under #[2,3]#.

This knowledge lets us put up the following equation of the integrals:
#int_2^3 f(x)\ dx=int_0^3 f(x)\ dx-int_0^2 f(x)\ dx#

Since we know the values for the integrals on the right hand side, we get:
#int_2^3 f(x)\ dx=6-5#

This means that the integral is equal to:
#int_2^3 f(x)\ dx=1#

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