What would you like to understand?

To evaluate the definite integral, substitute $x = 1 + e^u$, which gives $dx = e^u du$. Convert the bounds from $u=0$ and $u=\ln 2$ to $x=2$ and $x=3$.

Conditions: The integral is definite with bounds $u=0$ to $u=\ln 2$.; The integrand is $e^u \sqrt{1+e^u}$.; Substitution $x = 1 + e^u$ is used.

To evaluate the definite integral, substitute $x = 1 + e^u$, which gives $dx = e^u du$. Convert the bounds from $u=0$ and $u=\ln 2$ to $x=2$ and $x=3$.

Conditions: The integral is definite with bounds $u=0$ to $u=\ln 2$.; The integrand is $e^u \sqrt{1+e^u}$.; Substitution $x = 1 + e^u$ is used.

The substitution works because the derivative of $1+x^4$ is $4x^3$, which matches the $x^3$ term present in the integrand $x^3\sqrt{1+x^4}$. This allows the integral to be transformed into a simpler power function form, enabling the calculation of the antiderivative required for the integration by parts step.

Conditions: The term to be integrated is $x^3\sqrt{1+x^4}$.; The substitution variable is $w = 1+x^4$.; The differential $dw$ is proportional to $x^3 dx$.

The substitution works because the derivative of $1+x^4$ is $4x^3$, which matches the $x^3$ term present in the integrand $x^3\sqrt{1+x^4}$. This allows the integral to be transformed into a simpler power function form, enabling the calculation of the antiderivative required for the integration by parts step.

Conditions: The term to be integrated is $x^3\sqrt{1+x^4}$.; The substitution variable is $w = 1+x^4$.; The differential $dw$ is proportional to $x^3 dx$.

To set up the definite integral for the arc length of the curve $y = x^{3/2}$ over the interval $[0, 32/9]$, you first apply the power rule to find the derivative of the function, which is $f'(x) = \frac{3}{2}x^{1/2}$. Next, you square this derivative to get $(f'(x))^2 = \frac{9}{4}x$.

Conditions: The curve is defined by the function $f(x) = x^{3/2}$.; The interval of integration is $[0, 32/9]$.; The arc length formula $\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx$ is applicable.

To set up the definite integral for the arc length of the curve $y = x^{3/2}$ over the interval $[0, 32/9]$, you first apply the power rule to find the derivative of the function, which is $f'(x) = \frac{3}{2}x^{1/2}$. Next, you square this derivative to get $(f'(x))^2 = \frac{9}{4}x$.

Conditions: The curve is defined by the function $f(x) = x^{3/2}$.; The interval of integration is $[0, 32/9]$.; The arc length formula $\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx$ is applicable.

When using u-substitution in a definite integral, the original limits of integration in terms of $x$ must be converted into new limits in terms of $u$. This is done by substituting the original $x$-bounds into the substitution equation $u(x)$.

Conditions: The integral is a definite integral.; A substitution $u = u(x)$ is being used.; The original limits of integration are given in terms of $x$.

When using u-substitution in a definite integral, the original limits of integration in terms of $x$ must be converted into new limits in terms of $u$. This is done by substituting the original $x$-bounds into the substitution equation $u(x)$.

Conditions: The integral is a definite integral.; A substitution $u = u(x)$ is being used.; The original limits of integration are given in terms of $x$.