Question

Evaluate the following integrals. $$\int_{9}^{16} \sqrt{1+\sqrt{x}} d x$$

Short answer

After applying Gaussian substitution, power rule, and evaluating the antiderivative at the endpoints, the definite integral of the given function is approximately 213.173.

Step-by-step solution

Applying Gaussian substitution

We'll use the Gaussian substitution method to find the antiderivative of the given function. We will let \(u = \sqrt{x}\), which implies that \(x = u^2\). So, we will need to find the differential \(dx\) as a function of \(u\). Therefore, we have: $$\frac{d x}{d u} = 2 u \Rightarrow d x = 2 u d u$$ Now let's rewrite the integral in terms of \(u\): $$\int_{9}^{16} \sqrt{1+\sqrt{x}} d x = \int_{3}^{4} \sqrt{1+u}\cdot 2u du$$ Notice that the limits of integration have also changed.

Evaluating the integral in terms of \(u\)

Now we have the following integral: $$2 \int_{3}^{4} u\sqrt{u+1} du$$ Let's make a further substitution \(v = u+1\), then \(u = v - 1\) and we will remember to also change the limits of integration. When \(u = 3\), \(v = 4\) and when \(u = 4\), \(v = 5\). Therefore, the integral becomes: $$2 \int_{4}^{5} (v-1)\sqrt{v} dv$$ Now, we can apply the power rule and reverse the substitution.

Applying the power rule to the integral

Applying the power rule, we will first expand \((v-1)\sqrt{v}\) as \(v\sqrt{v}-\sqrt{v}\), which simplifies further to: \(v^{\frac{3}{2}} - v^{\frac{1}{2}}\). Now we can integrate term by term: $$2 \left( \int_{4}^{5} v^{\frac{3}{2}} dv - \int_{4}^{5} v^{\frac{1}{2}} dv \right)$$ Integrating the first term: $$\frac{2}{\frac{5}{2}}\left[ v^{\frac{5}{2}}\right]_{4}^{5} = \frac{4}{5}\left[ v^{\frac{5}{2}}\right]_{4}^{5}$$ Integrating the second term: $$2\left[ 2v^{\frac{3}{2}} \right]_{4}^{5} = 4\left[ v^{\frac{3}{2}} \right]_{4}^{5}$$

Finding the difference of the antiderivative at the endpoints

Now we simply need to subtract the antiderivative evaluated at the lower endpoint from the antiderivative evaluated at the upper endpoint for the definite integral: First term: $$\frac{4}{5}\left[5^{\frac{5}{2}}-4^{\frac{5}{2}}\right]$$ Second term: $$-4\left[5^{\frac{3}{2}}-4^{\frac{3}{2}}\right]$$ Put them together: $$\int_{9}^{16} \sqrt{1+\sqrt{x}} d x = 2 \int_{3}^{4} u\sqrt{u+1} du = \frac{4}{5}\left[5^{\frac{5}{2}}-4^{\frac{5}{2}}\right] -4\left[5^{\frac{3}{2}}-4^{\frac{3}{2}}\right]$$ To finish, just evaluate the resulting expression to find the final answer: $$\frac{4}{5}(3125\sqrt{5}-512\sqrt{4}) -4(125\sqrt{5}-64\sqrt{4}) \approx 213.173$$

Key concepts

Gaussian Substitution

Gaussian substitution is a technique used to simplify integration by changing the variable to make the integral easier to solve. In this exercise, the method began with the substitution \( u = \sqrt{x} \). This means \( x = u^2 \), and the differential \( dx \) can be expressed in terms of \( du \) as \( dx = 2u \, du \). This change simplifies the original integral and adjusts the limits of integration from \( x \) values (9 to 16) to \( u \) values (3 to 4).

The purpose of this substitution is to transform a complicated function into one that can be handled more easily. After substitution, the integral becomes \( 2 \int_{3}^{4} u\sqrt{u+1} \, du \). Making such substitution is particularly helpful when dealing with square roots or other algebraic transformations.

Definite Integral

A definite integral calculates the accumulation of quantities, in this case over a specific interval. In this example, the interval is from 9 to 16. The definite integral of a function gives the net area between the function and the x-axis over this interval.

When using substitutions like in this example, be careful to adjust the limits of integration accordingly. After substituting \( u = \sqrt{x} \), the new limits are from 3 to 4 (since when \( x = 9 \), \( u = 3 \) and when \( x = 16 \), \( u = 4 \)). This allows the entire computation to solve in a new transformed variable setup.

Antiderivative

An antiderivative of a function is essentially its reverse derivative, representing a function whose derivative gives the original function. In integration, finding the antiderivative allows us to determine the area under the curve of a given function.

While working through our specific integral, substitutions transformed the function, making it easier to find an antiderivative. This involves algebraic simplifications and applying integration rules to find terms that reflect the original function when differentiated. In our step-by-step solution, the antiderivative was used to find the value of the original definite integral.

Power Rule

The power rule for integration is a basic technique that involves integrating functions of the form \( x^n \). The rule states:\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]where \( n eq -1 \). In this exercise, the power rule was used to simplify terms like \( v^{\frac{3}{2}} \) and \( v^{\frac{1}{2}} \).

Applying the power rule gives us these antiderivatives, allowing us to individually integrate terms within the integral. Remember to adjust these terms in terms of their respective limits once integrated, a crucial step for computing the definite integral. In this step, the constants and negative signs were carefully considered to find the accurate result for each term.