How do you find the definite integral for: #((2x)/(1+x^2)^(1/3))dx# for the intervals #[0, sqrt7]#?

Question

Paisley Johnson · Accepted Answer

First, find the indefinite integral and then evaluate from the lower bound to the upper bound.

#int_0^sqrt(7) (2x)/(1 + x^2)^(1/3)dx = ?# Let #u = 1 + x^2#, then #du = 2xdx# and: #int(2x)/(1 + x^2)^(1/3)dx = intu^(-1/3)du# #int(2x)/(1 + x^2)^(1/3)dx = 3/2u^(2/3) + C# Reverse the substitution: #int(2x)/(1 + x^2)^(1/3)dx = 3/2(1 + x^2)^(2/3) + C# We do not need the constant for the definite integral: #int_0^sqrt(7) (2x)/(1 + x^2)^(1/3)dx = 3/2(1 + x^2)^(2/3)|_0^sqrt(7)# #int_0^sqrt(7) (2x)/(1 + x^2)^(1/3)dx = 3/2(1 + (sqrt(7))^2)^(2/3) - 3/2(1 + 0^2)^(2/3)# #int_0^sqrt(7) (2x)/(1 + x^2)^(1/3)dx = 3/2(8)^(2/3) - 3/2# #int_0^sqrt(7) (2x)/(1 + x^2)^(1/3)dx = 4.5#

Charles Arocha · Answer

undefined To find the definite integral of $\frac{2x}{(1+x^2)^{1/3}}$ over the interval $[0, \sqrt{7}]$, you integrate the function with respect to $x$ and evaluate the integral at the upper and lower limits of integration.

The antiderivative of $\frac{2x}{(1+x^2)^{1/3}}$ is $(1+x^2)^{2/3}$.

Evaluate the antiderivative at the upper and lower limits of integration:

At $x = \sqrt{7}$, the antiderivative is $(1+(\sqrt{7})^2)^{2/3} = 8^{1/3} = 2$.

At $x = 0$, the antiderivative is $(1+0^2)^{2/3} = 1$.

Subtract the value of the antiderivative at the lower limit from the value at the upper limit:

$2 - 1 = 1$.

Therefore, the definite integral of $\frac{2x}{(1+x^2)^{1/3}}$ over the interval $[0, \sqrt{7}]$ is $1$.

Paisley Johnson · Answer

undefined To find the definite integral of ((2x)/(1+x^2)^(1/3))dx for the intervals [0, sqrt7], you can use the substitution method. Let u = 1 + x^2. Then, du = 2x dx.

After substituting and simplifying, you'll have an integral in terms of u.

Then, find the antiderivative of the expression in terms of u.

Once you find the antiderivative, evaluate it at the upper and lower limits of integration, and subtract the value at the lower limit from the value at the upper limit to find the definite integral.