/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Evaluate the definite integral. ... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the definite integral. $$\int_{0}^{1} \sqrt{2 x}(\sqrt{x}+\sqrt{2}) d x$$

Short Answer

Expert verified
The definite integral is: \[\int_{0}^{1} \sqrt{2 x}(\sqrt{x}+\sqrt{2}) d x = \frac{4}{5} + \frac{4\sqrt{2}}{3}.\]

Step by step solution

01

Rewrite the integrand with a simple term.

To make the function easier to work with, we can rewrite the integrand by expanding the square root terms and grouping the constants: \[\sqrt{2x}(\sqrt{x}+\sqrt{2}) = \sqrt{x}(2x + 2\sqrt{2}).\]
02

Simplify the integrand further.

Now, let's simplify the function by distributing the square root term and then rewrite the expression under the square roots as powers: \[2x\sqrt{x} + 2\sqrt{2x} = 2x^{\frac{3}{2}} + 2\sqrt{2}x^{\frac{1}{2}}.\]
03

Find the antiderivative of the integrand.

Now, we have to find the antiderivative of the simplified function. When integrating, we use the power rule, which means adding 1 to the exponent and dividing by the new exponent: \[\int(2x^{\frac{3}{2}} + 2\sqrt{2}x^{\frac{1}{2}})dx = \frac{4}{5}x^{\frac{5}{2}} + \frac{4\sqrt{2}}{3}x^{\frac{3}{2}} + C.\] The constant of integration (C) can be ignored when evaluating definite integrals.
04

Evaluate the integral using the limits of integration.

Now we need to evaluate the definite integral with the given limits 0 and 1. To do this, we first plug in the upper limit, and then subtract the result of the lower limit: \[\left[\frac{4}{5}x^{\frac{5}{2}} + \frac{4\sqrt{2}}{3}x^{\frac{3}{2}}\right]_0^1 = \left(\frac{4}{5}(1)^{\frac{5}{2}} + \frac{4\sqrt{2}}{3}(1)^{\frac{3}{2}}\right) - \left(\frac{4}{5}(0)^{\frac{5}{2}} + \frac{4\sqrt{2}}{3}(0)^{\frac{3}{2}}\right).\]
05

Calculate the definite integral value.

Let's simplify the equation to find the value of the definite integral: \[\left(\frac{4}{5} + \frac{4\sqrt{2}}{3}\right) - \left(0 + 0\right) = \frac{4}{5} + \frac{4\sqrt{2}}{3}.\] Thus, the definite integral is: $$\int_{0}^{1} \sqrt{2 x}(\sqrt{x}+\sqrt{2}) d x = \frac{4}{5} + \frac{4\sqrt{2}}{3}.$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Antiderivative
An antiderivative of a function is another function that reverses the process of differentiation. Essentially, when you find an antiderivative, you're seeking a function whose derivative is the original function you began with.
Let's think of it as finding the original function hidden behind the curtain of differentiation.
  • The process of finding an antiderivative is called integration.
  • Antiderivatives include a constant of integration, represented by the symbol \(C\), since the differentiation of any constant is zero.
When working with definite integrals, we turn our focus to finding the antiderivative and then using the limits of integration to evaluate the entire integral. In these cases, the constant \(C\) is canceled out because it affects both the upper and lower limits equally. This was demonstrated in our example when we found the antiderivative:\[ \int(2x^{\frac{3}{2}} + 2\sqrt{2}x^{\frac{1}{2}})dx = \frac{4}{5}x^{\frac{5}{2}} + \frac{4\sqrt{2}}{3}x^{\frac{3}{2}} + C. \]
Power Rule
The power rule is a fundamental technique used in calculus to find the derivative or antiderivative of functions with the form \(x^n\). It's a tool that simplifies the integration process significantly, especially when dealing with polynomials or power functions.
To use the power rule in integration:
  • Add one to the exponent \(n\).
  • Divide by the new exponent.
The general expression for the power rule in integration is:\[\int{x^n}dx = \frac{x^{n+1}}{n+1} + C, \quad \text{where} \ n eq -1.\]In the exercise, the power rule was applied to each term of the function \(2x^{\frac{3}{2}} + 2\sqrt{2}x^{\frac{1}{2}}\). By adding 1 to the exponent and then dividing by this new power, we make the integral solvable, arriving at the expression:\[\frac{4}{5}x^{\frac{5}{2}} + \frac{4\sqrt{2}}{3}x^{\frac{3}{2}}\]
Limits of Integration
The concept of limits of integration is unique to definite integrals, allowing us to calculate an exact numerical value for the area under a curve between two specified points. These limits essentially "slice" our domain into a start and endpoint.
In practice, limits of integration are represented as numbers at the top (upper limit) and bottom (lower limit) of the integral symbol and help define where the area computation begins and ends.
  • The upper limit indicates where to stop evaluating the antiderivative.
  • The lower limit marks where to start.
To evaluate using these limits, plug the upper limit into the antiderivative, subtract the result of the lower limit, and voilà—a definite answer emerges. This was visible in the step-by-step earlier as each integral computation:\[\left[\frac{4}{5}x^{\frac{5}{2}} + \frac{4\sqrt{2}}{3}x^{\frac{3}{2}}\right]_0^1 = \left(\frac{4}{5}(1)^\frac{5}{2} + \frac{4\sqrt{2}}{3}(1)^\frac{3}{2}\right) - \left(\frac{4}{5}(0)^\frac{5}{2} + \frac{4\sqrt{2}}{3}(0)^\frac{3}{2}\right)\]Using the limits allowed us to compute the exact value of the definite integral, giving our sought result.

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Most popular questions from this chapter

During the construction of ? high-rise apartment building, a construction worker accidently drops a hammer that falls vertically a distance of \(h \mathrm{ft}\). The velocity of the hammer after falling a distance of \(x \mathrm{ft}\) is \(v=\sqrt{2 g x} \mathrm{ft} / \mathrm{sec}(0 \leq x \leq h) .\) Show that the aver- age velocity of the hammer over this path is \(\bar{v}=\frac{2}{3} \sqrt{2 g h}\) \(\mathrm{ft} / \mathrm{sec} .\)

Sketch the graphs of the functions \(f\) and \(g\) and find the area of the region enclosed by these graphs and the vertical lines \(x=a\) and \(x=b\). $$f(x)=\frac{1}{x}, g(x)=2 x-1 ; a=1, b=4$$

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. \(\int_{0}^{1} x \sqrt{x+1} d x=\sqrt{x+1} \int_{0}^{1} x d x=\left.\frac{1}{2} x^{2} \sqrt{x+1}\right|_{0} ^{1}=\frac{\sqrt{2}}{2}\)78. If \(f^{\prime}\) is continuous on \([0,2]\), then \(\int_{0}^{2} f^{\prime}(x) d x=f(2)-f(0)\).

Verify by direct computation that \(\int_{0}^{3}\left(1+x^{3}\right) d x\) \(\quad=\int_{0}^{1}\left(1+x^{3}\right) d x+\int_{1}^{2}\left(1+x^{3}\right) d x+\int_{2}^{3}\left(1+x^{3}\right) d x\) hence showing that Property 5 may be extended.

Show that the area of a region \(R\) bounded above by the graph of a function \(f\) and below by the graph of a function \(g\) from \(x=a\) to \(x=b\) is given by $$ \int_{a}^{b}[f(x)-g(x)] d x $$ Hint: The validity of the formula was verified earlier for the case when both \(f\) and \(g\) were nonnegative. Now, let \(f\) and \(g\) be two functions such that \(f(x) \geq g(x)\) for \(a \leq x \leq b\). Then, there exists some nonnegative constant \(c\) such that the curves \(y=f(x)+c\) and \(\bar{y}=\) \(g(x)+c\) are translated in the \(y\) -direction in such a way that the region \(R^{r}\) has the same area as the region \(R\) (see the accompanying figures). Show that the area of \(R^{\prime}\) is given by $$ \int_{a}^{b}\\{[f(x)+c]-[g(x)+c]\\} d x=\int_{a}^{b}[f(x)-g(x)] d x $$

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