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Chapter 7: Problem 44

Use the approaches discussed in this section to evaluate the following integrals. $$\int_{0}^{1} \sqrt{1+\sqrt{x}} d x$$

Short Answer

Expert verified
Question: Evaluate the definite integral of the function $$f(x) = \sqrt{1+\sqrt{x}}$$ from 0 to 1. Answer: The value of the integral is $$\int_{0}^{1} \sqrt{1+\sqrt{x}} d x= \frac{24\sqrt{2}-72}{15}$$

Step by step solution

01

Identify the function to be evaluated

The function to be evaluated is: $$f(x) = \sqrt{1+\sqrt{x}}$$ Evaluate the definite integral of this function from 0 to 1: $$\int_{0}^{1} \sqrt{1+\sqrt{x}} d x$$
02

Simplify the function using substitution

Let $$u = \sqrt{x},$$ so $$x = u^2$$. Therefore, $$d x = 2 u d u$$. Now we can rewrite the integral in terms of $$u$$: $$ \int \sqrt{1+u} * 2u d u$$
03

Adjust the limits of integration

The original limits of integration were given in terms of $$x$$ (0 and 1). We need to adjust them to be in terms of $$u$$. When $$x=0, u=\sqrt{0}=0$$, and when $$x=1, u=\sqrt{1}=1$$. So, the new limits of integration are still 0 and 1: $$ \int_{0}^{1} 2u\sqrt{1+u} d u$$
04

Evaluate the integral

Now, let's evaluate the integral: $$\int_{0}^{1} 2u\sqrt{1+u} d u$$ Let's make another substitution: $$v = 1+u$$, so $$u = v-1$$ and $$d u = d v$$. This gives us: $$ \int 2(v-1)\sqrt{v} d v$$ And the limits of integration become: When $$u=0, v=0+1=1$$ and when $$u=1, v=1+1=2$$ Now, the integral is: $$ \int_{1}^{2} 2(v-1)\sqrt{v} d v$$
05

Break down the integral

Distribute the $$2$$: $$ \int_{1}^{2} (2v - 2)\sqrt{v} d v$$ Break the integral into two parts: $$2\int_{1}^{2} v\sqrt{v} d v - 2\int_{1}^{2} \sqrt{v} d v$$
06

Evaluate each part of the integral using antiderivatives

For the first integral, use power rule (add 1 to the exponent and divide by the new exponent): $$2\int_{1}^{2} v^{3/2} d v = \frac{4}{5}v^{5/2}\Big|_1^2$$ For the second integral, do the same: $$-2\int_{1}^{2}v^{1/2} d v=-\frac{4}{3}v^{3/2}\Big|_1^2$$
07

Combine the results and evaluate the definite integral

Combine the two results: $$\frac{4}{5}v^{5/2}-\frac{4}{3}v^{3/2}\Big|_1^2$$ Evaluate using the limits of integration: $$\left(\frac{4}{5}(2^{5/2})-\frac{4}{3}(2^{3/2})\right)-\left(\frac{4}{5}(1^{5/2})-\frac{4}{3}(1^{3/2})\right)$$ Simplify and calculate the result: $$\left(\frac{8\sqrt{2}}{5}-\frac{8}{3}\right)-\left(\frac{4}{5}-\frac{4}{3}\right)$$ $$= \left(\frac{3\cdot8\sqrt{2}-16\cdot5}{15} \right)-\left(\frac{4\cdot3-16}{15}\right)$$ $$= \frac{24\sqrt{2}-80}{15}-\frac{-8}{15}= \frac{24\sqrt{2}-72}{15}$$ So, the value of the integral is: $$\int_{0}^{1} \sqrt{1+\sqrt{x}} d x= \frac{24\sqrt{2}-72}{15}$$

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Most popular questions from this chapter

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