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

2 Answers
Jun 10, 2017

#16.5#

Explanation:

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#

Jun 10, 2017

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

Explanation:

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#