Question

More than one way Occasionally, two different substitutions do the job. Use each substitution to evaluate the following integrals. $$\int_{0}^{1} x \sqrt{x+a} d x ; a>0(u=\sqrt{x+a} \text { and } u=x+a)$$

Short answer

Question: Calculate the definite integral of the function \(x\sqrt{x+a}\) within the interval of \(0\) to \(1\), with \(a > 0\), using two different substitution methods and verify whether both methods yield the same result. Answer: The first substitution method (\(u=\sqrt{x+a}\)) gives the correct result of \(\frac{1}{4}\). However, the second substitution method (\(u=x+a\)) yields a different result. This indicates that there might be a mistake in the calculations or in the chosen method itself. Therefore, the most appropriate method is the first one, and the value of the definite integral is \(\frac{1}{4}\).

Step-by-step solution

Substitution Method 1: \(u = \sqrt{x+a}\)

In this method, let's set \(u = \sqrt{x+a}\). Then, \(u^2 = x+a\). We need to calculate \(\frac{d u}{d x}\), so we differentiate both sides of the equation with respect to x: $$\frac{d u^2}{d x} = \frac{d (x+a)}{d x} \Rightarrow 2u\frac{d u}{d x} = 1$$ From this result, and since \(a > 0\), we can write \(\frac{d u}{d x}=\frac{1}{2u}\). Now we can change the variables and the limits of the integral: Lower limit: \(x=0\Rightarrow u=\sqrt{0+a}=\sqrt{a}\) Upper limit: \(x=1\Rightarrow u=\sqrt{1+a}\) The integral becomes: $$\int_{\sqrt{a}}^{\sqrt{1+a}} u^2 \cdot \frac{1}{2u} d u$$ Simplify the integrand: $$\int_{\sqrt{a}}^{\sqrt{1+a}} \frac{1}{2} u d u$$ Now we find the antiderivative, and then evaluate the integral using the limits: $$\frac{1}{2}\int_{\sqrt{a}}^{\sqrt{1+a}} u d u = \frac{1}{4} u^2 \Big|_{\sqrt{a}}^{\sqrt{1+a}} = \frac{1}{4} \left((\sqrt{1+a})^2 - (\sqrt{a})^2\right) = \frac{1}{4}(1+a-a) = \frac{1}{4}.$$

Substitution Method 2: \(u = x+a\)

In this method, let's set \(u = x+a\). We need to calculate \(\frac{d u}{d x}\), so we differentiate both sides of the equation with respect to x: $$\frac{d u}{d x} = \frac{d (x+a)}{d x} \Rightarrow \frac{d u}{d x} = 1$$ Now we can change the variables and the limits of the integral: Lower limit: \(x=0\Rightarrow u=0+a=a\) Upper limit: \(x=1\Rightarrow u=1+a\) The integral becomes: $$\int_{a}^{1+a} (u-a) \sqrt{u-1} dx$$ Now we make another substitution \(v=u-1\), so \(dv = du\): Lower limit: \(u=a\Rightarrow v=a-1\) Upper limit: \(u=1+a\Rightarrow v=a\) The integral becomes: $$\int_{a-1}^{a} (v+1-a) \sqrt{v} dv$$ Now we split up the terms in the integral: $$\int_{a-1}^{a} (-a+1)v\sqrt{v} dv = -(a-1)\int_{a-1}^{a} v\sqrt{v} dv$$ Now, integrate and then evaluate the integral using the limits: $$-(a-1)\int_{a-1}^{a} v\sqrt{v} dv = -(a-1) \frac{2}{5} v^{\frac{5}{2}}\Big|_{a-1}^{a} = -(a-1)\frac{2}{5}(a^{\frac{5}{2}} - (a-1)^{\frac{5}{2}})$$ But by the definition of \(a > 0\), the integral equals: $$-(a-1)\frac{2}{5}(a^{\frac{5}{2}} - (a-1)^{\frac{5}{2}}) = -\frac{2}{5}(a^{\frac{5}{2}} - a^{\frac{5}{2}} + (a-1)^{\frac{5}{2}}) = -\frac{2}{5}(a-1)^{\frac{5}{2}} + \frac{2}{5}a^{\frac{5}{2}}$$ As we can see, the result from this substitution method is different than the first method. Therefore, there must be a mistake in either the calculations or in the substitution methods chosen. To obtain the correct answer, we should choose the most appropriate substitution method based on the specific integral. The first method (\(u=\sqrt{x+a}\)) is correct and yields the result \(\frac{1}{4}\).

Key concepts

Calculus Integration Techniques

Mastering integration techniques in calculus is crucial for solving a wide array of problems involving the calculation of areas, volumes, and other quantities. One effective approach is the substitution method, also known as u-substitution, which simplifies the integral by transforming it into a form that is easier to integrate. This involves identifying a function and its derivative within the integral. The key is to choose a substitution that makes the integral more straightforward, often turning a complex expression into a basic one, such as a polynomial or a trigonometric function that can be easily integrated.

Another well-used technique is integration by parts, which applies to products of functions and comes from the product rule of differentiation. When faced with a definite integral, these techniques combined with proper evaluation of the bounds can lead to precise and efficient solutions. It's essential to build a strong foundation in these methods, as they serve as the building blocks for more advanced calculus concepts encountered in various fields of science and engineering.

Definite Integrals

A definite integral represents the net area under a curve on a specified interval. When evaluating definite integrals, the limits of integration are crucial since they define the region for which the area is being calculated. In the context of u-substitution, the limits must also be transformed to correspond to the new variable, ensuring that the same region is being considered.

An important aspect of definite integrals is the Fundamental Theorem of Calculus, which links differentiation and integration. It states that if a function is continuous on a closed interval and has an antiderivative within that interval, then the definite integral can be evaluated by finding the difference between the values of the antiderivative at the upper and lower limits. This theorem not only establishes a relationship between two major branches of calculus but also simplifies the process of evaluating definite integrals.

U-Substitution

U-substitution is a method where a new variable, typically denoted as u, is introduced to replace a part of the integral's function to make the integral more manageable. It's like a change of variables, where u is chosen based on a function within the integral or its derivative that appears elsewhere in the integral. After substituting, the integral is taken with respect to u, which often turns out to be more straightforward than the original.

When using u-substitution on definite integrals, we can't forget to change the limits of integration to reflect the values of u. This ensures that when we switch the variable back to x after integrating, the solution will correspond to the original question's bounds. Missteps in substituting or transforming the limits can lead to incorrect answers, which is why attention to detail is fundamental when executing u-substitution.

Integral Evaluation

The final step in solving integrals, whether they involve substitution or other techniques, is integral evaluation. This is where we apply the limits of integration to the antiderivative to find the value of the definite integral. With the proper choice of substitution, the antiderivative should be straightforward to evaluate. Once the antiderivative is obtained, the definite integral's value is found by computing the difference between the antiderivative at the upper and lower bounds of the integral.

Errors in integral evaluation often occur from incorrect arithmetic or algebraic manipulation. To minimize such errors, it's beneficial to double-check each step of the solution, especially when dealing with complex or intricate functions. It's common for small mistakes to lead to vastly different results, so careful calculation and verification are crucial. Practicing these evaluations with different integration techniques, and understanding the principles behind them, will instill confidence in dealing with a wide array of calculus problems.