How to solve definite integrals (when you have to change bounds)

The new integration bounds are $x=2$ and $x=3$. When $u=0$, $x = 1 + e^0 = 1 + 1 = 2$.

Conditions: The substitution is $x = 1 + e^u$.; The original bounds are $u=0$ and $u=\ln 2$.; The exponential function $e^u$ is evaluated at these points.

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 $x = 1 + e^u$ simplifies the integrand because its derivative $dx = e^u du$ exactly matches the $e^u du$ term in the original integral. This allows the complex expression $e^u \sqrt{1+e^u} du$ to be rewritten as the simpler radical $\sqrt{x} dx$.

Conditions: The integrand contains $e^u \sqrt{1+e^u}$.; The substitution is $x = 1 + e^u$.; The differential is $dx = e^u du$.

The antiderivative of $\sqrt{x}$ is $\frac{2}{3}x^{3/2}$. This is found by rewriting $\sqrt{x}$ as $x^{1/2}$ and applying the power rule for integration, which states that $\int x^n dx = \frac{x^{n+1}}{n+1}$ for $n \neq -1$.

Conditions: The function is $\sqrt{x}$.; The power rule for integration is applied.; The exponent $n = 1/2$.

The bounds must be converted because the variable of integration changes from $u$ to $x$. The original bounds correspond to values of $u$, but the new integral is evaluated with respect to $x$.

Conditions: The integral is a definite integral.; A substitution $x = g(u)$ is performed.; The variable of integration changes from $u$ to $x$.

The derivative of $1 + e^u$ is $e^u$, which appears as a factor in the integrand $e^u \sqrt{1+e^u}$. This allows the entire term $e^u du$ to be replaced by $dx$, simplifying the integral to $\int \sqrt{x} dx$.

Conditions: The integrand is $e^u \sqrt{1+e^u}$.; The substitution is $x = 1 + e^u$.; The derivative $dx/du = e^u$ is present in the integrand.

The final value is $2\sqrt{3} - \frac{4\sqrt{2}}{3}$. This is obtained by evaluating the antiderivative $\frac{2}{3}x^{3/2}$ at the converted bounds $x=3$ and $x=2$ and subtracting the results.

Conditions: The integral is $\int_0^{\ln 2} e^u \sqrt{1+e^u} du$.; The substitution $x = 1 + e^u$ is used.; The bounds are converted to 2 and 3.; The antiderivative is $\frac{2}{3}x^{3/2}$.

To evaluate $\int_2^3 \sqrt{x} dx$, first find the antiderivative $\frac{2}{3}x^{3/2}$. Then, apply the Fundamental Theorem of Calculus by subtracting the value of the antiderivative at the lower bound from its value at the upper bound: $\left[\frac{2}{3}x^{3/2}\right]_2^3 = \frac{2}{3}(3)^{3/2} - \frac{2}{3}(2)^{3/2}$.

Conditions: The integral is $\int_2^3 \sqrt{x} dx$.; The antiderivative is $\frac{2}{3}x^{3/2}$.; The bounds are 2 and 3.