How do you evaluate the indefinite integral #int (x^2-x+5)dx#?

Question

4687 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-22T22:27:45

#int(x^2-x+5)dx=x^3/3-x^2/2+5x+C#

Remember that an indefinite integral is an antiderivative, and since the derivative of sums is the sum of derivatives then

#intf(x)+g(x)+h(x)dx=intf(x)dx+intg(x)dx+inth(x)dx#

it is the same for integrals

In this case

#f(x)=x^2#

#g(x)=-x#

#h(x)=5#

Since #intx^ndx=x^(n+1)/(n+1)#

#intf(x)dx=intx^2dx=x^3/3#

and also since #int-xdx=-intxdx#

#intg(x)dx=int(-x)dx=-intxdx=-x^2/2#

and since for a constant #c#, #intcdx=cx# then

#inth(x)dx=int5dx=5x#

Then put them together

#int(x^2-x+5)dx=x^3/3-x^2/2+5x+C#

Answer 2 · 2017-01-22T22:29:14

I tried this:

We can break it into three parts and write:
#intx^2dx-intxdx+int5dx=#

we now use the general integration formula as:
#color(red)(intx^ndx=x^(n+1)/(n+1)+c)#

we can write our integral as:
#intx^2dx-intxdx+5intx^0dx=#
and we get:
#=x^3/3-x^2/2+5x+c#