How do you find the definite integral for: #((2x^(2) +3)dx# for the intervals #[4, 6]#?

Question

Elijah Abram · Accepted Answer

#int_4^6(2x^2+3)dx=322/3#

The definite integral of #2x^2+3# on the interval #[4,6]# can be expressed as:

#int_4^6(2x^2+3)dx#

To evaluate this, find the antiderivative #F(x)# of #f(x)=2x^2+3# and then find #F(6)-F(4)#.

The antiderivative can be found using the rule:

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

So, we can express the definite integral as:

#int_4^6(2x^2+3)dx=2int_4^6x^2+3int_4^6dx#

Note that #3intdx=3x#, because #intdx# can be thought of as #intx^0dx=x^(0+1)/(0+1)=x#.

#2int_4^6x^2+3int_4^6dx=[2(x^(2+1)/(2+1))+3x]_4^6#

#=[2/3x^3+3x]_4^6#

Note that this is saying the same thing as #F(6)-F(4)#, where #F(x)# is the antiderivative #F(x)=2/3x^3+3x#.

#[2/3x^3+3x]_4^6=(2/3(6)^3+6(3))-(2/3(4)^3+4(3))#

#=(2/3(216)+18)-(2/3(64)+12)#

#=(432/3+54/3)-(128/3+36/3)#

#=(432+54-128-36)/3#

#=322/3#

This can be interpreted to mean that the area bounded by the curve #2x^2+3# on #4 < x < 6# is #322/3#.

Ezekiel Alexander · Answer

undefined To find the definite integral of $ (2x^2 + 3) $ over the interval $[4, 6]$, follow these steps:

1. Find the antiderivative of $2x^2 + 3$ with respect to $x$.
2. Evaluate the antiderivative at the upper limit of integration (6).
3. Evaluate the antiderivative at the lower limit of integration (4).
4. Subtract the result from step 3 from the result from step 2.

Let's proceed with these steps:

1. The antiderivative of $2x^2 + 3$ with respect to $x$ is $\frac{2}{3}x^3 + 3x + C$, where $C$ is the constant of integration.
2. Evaluating the antiderivative at $x = 6$, we get $\left(\frac{2}{3}(6)^3 + 3(6)\right) = 144 + 18 = 162$.
3. Evaluating the antiderivative at $x = 4$, we get $\left(\frac{2}{3}(4)^3 + 3(4)\right) = 85.\overline{3} + 12 = 97.\overline{3}$.
4. Subtracting the result from step 3 from the result from step 2, we get $162 - 97.\overline{3} = 64.\overline{6}$.

Therefore, the definite integral of $ (2x^2 + 3) $ over the interval $[4, 6]$ is $64.\overline{6}$.