How do you evaluate the definite integral #int (2x^3)dx# from [1,3]?

Question

Paisley Johnson · Accepted Answer

#int_1^3 2x^3dx = 40#

We will start by thinking about the integral as I with the limits:

#I = int_1^3 2x^3dx#

The constant will be removed from the multiplication with the variable as it is:

#I = 2 int_1^3 x^3dx#

We are aware of the integration power rule:

#I = int x^ndx = (x^(n+1))/(n+1)#

Using the integral and the power rule

#I = 2 [(x^(3+1))/(3+1)]_1^3#

#I = 2 [(x^4)/4]_1^3#

We can now set the integration's boundaries because we are aware of the following rule:

Higher and Lower Boundaries

Hence:

#I = 2 [((3^4)/4)-(1^4/4)]#

We know, #3^4=81# and #1^4=1#

#I = 2 [((81)/4)-(1/4)]#

As the base is same, we can directly subtract #81# and #1#

#I = 2 [((81-1)/4)]#

#I = 2 [80/4]#

We can divide #80# by #4#

#I = 2.20#

#I=40#

Hence:

#int_1^3 2x^3dx = 40#

Christopher Agresta · Answer

undefined To evaluate the definite integral of $ \int_{1}^{3} 2x^3 \, dx $, we first find the antiderivative of the integrand. In this case, the antiderivative of $ 2x^3 $ is $ \frac{1}{2}x^4 $. Then, we apply the Fundamental Theorem of Calculus, which states that if $ F(x) $ is an antiderivative of $ f(x) $ on an interval $[a, b]$, then $\int_{a}^{b} f(x) \, dx = F(b) - F(a)$.

So, $ \int_{1}^{3} 2x^3 \, dx = \left[ \frac{1}{2}x^4 \right]_{1}^{3} = \frac{1}{2}(3^4) - \frac{1}{2}(1^4) = \frac{81}{2} - \frac{1}{2} = \frac{80}{2} = 40 $.

Therefore, the value of the definite integral is 40.