How do you evaluate the definite integral #int (2x) dx# from #[2,3]#?

Question

Emma Aldridge · Accepted Answer

#5#

Use the Power Rule for integration when dealing with the indefinite integral:

#int x^n \ dx = (x^(n+1))/(n+1) + C#

And, with constant #alpha#:

#int alpha x^n \ dx = (alpha \ x^(n+1))/(n+1) + C#

Alternately, lift the constant outside of the integration if you'd like:

#int alpha x^n \ dx = alpha int x^n \ dx#

#=alpha ( \ x^(n+1))/(n+1) + C = (alpha \ x^(n+1))/(n+1) + C#

I'm purposefully pushing this.

So

#int 2 x \ dx#

# = 2 int x^color(red)(1) \ dx#

from the Power Rule # =2 ( x^(1+1))/(1+1) + C#

# =x^2 + C qquad triangle#

Finally, if in doubt, differentiate your result in #triangle#, because differentiation and integration are like inverse processes

# d/dx (x^2 + C) = d/dx (x^2) + d/dx(C) = 2x + 0 = 2x# Voila!!

The definite integral is next.

#int_2^3 2 x \ dx#

#= 2 int_2^3 x \ dx#

from the Power Rule #= 2 [ x^2/2 ]_2^3#

#= [ x^2 ]_2^3#

#= 9 - 4 = 5#

Scarlett Allison · Answer

undefined To evaluate the definite integral of $ \int_{2}^{3} (2x) \, dx $, you use the fundamental theorem of calculus. First, you integrate $2x$ with respect to $x$, which gives you $x^2$. Then, you evaluate this antiderivative at the upper and lower bounds of integration and find the difference.

$$ \int_{2}^{3} (2x) \, dx = [x^2]_{2}^{3} = (3^2) - (2^2) = 9 - 4 = 5 $$

So, the value of the definite integral is 5.