How do you find #\int _ { 0} ^ { 3} ( x ^ { 2} + x + 1) d x#?

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Answer 1 · 2017-06-10T19:42:45

#16.5#

Let's take apart the integral.

#int (x^2+x+1) dx = intx^2 dx + int x dx + int 1 dx#
#=x^3/3+x^2/2+x+C#

Then, let's substitute the upper and lower bounds back in.

#3^3/3+3^2/2+3-0#
#=9+9/2+3=16.5#

Answer 2 · 2017-06-10T23:55:27

# int_0^3 \ x^2+x+1 \ dx = 33/2#

Integrating each term, we have:

# int_0^3 \ x^2+x+1 \ dx = [x^3/3+x^2/2+x]_0^3#
# " " = (9+9/2+3) - (0)#2
# " " = 33/2#