/*! 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 5 Evaluate \(\int_{0}^{1}\left(1+5... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 5

Evaluate \(\int_{0}^{1}\left(1+5 x-x^{5}\right)^{4}\left(x^{2}-1\right)\left(x^{2}+1\right) d x\)

Short Answer

Expert verified
Question: Evaluate the definite integral of the given function: $$\int_{0}^{1}(1+5x-x^5)^4(x^4 - 1) dx$$ Answer: The definite integral of the given function is 0.

Step by step solution

01

Expand the integrand

Distribute (1+5x-x^5)^4 and multiply by (x^2-1)(x^2+1). However, since this distributive step will be quite tedious and may cause confusion, we can look for a simpler way to compute the integral. Notice that the product (x^2-1)(x^2+1) = x^4 - 1. Now our integral has the form: $$\int_{0}^{1}(1+5x-x^5)^4(x^4 - 1) dx$$ Step 2: Integrate term by term
02

Use power rule, constant factor rule, and sum/difference rule

Apply the power rule, constant factor rule, and sum/difference rule to integrate term by term, treating (1+5x-x^5)^4 as a constant factor: $$\int_{0}^{1}(1+5x-x^5)^4(x^4 - 1) dx = \int_{0}^{1}(1+5x-x^5)^4 x^4 dx - \int_{0}^{1}(1+5x-x^5)^4 dx$$ Now integrate $$\int_{0}^{1}(1+5x-x^5)^4 x^4 dx$$ and $$\int_{0}^{1}(1+5x-x^5)^4 dx$$ separately. Step 3: Evaluate the definite integrals
03

Apply the power rule and integrate

For simplifying the integrals, substitution method is used Let u = 1+5x-x^5. Then, du/dx = -5*x^4 + 5, so the first integral becomes $$\int_{0}^{1}(1+5x-x^5)^4 x^4 dx = \frac{1}{-20}\int_{1}^{1}(u)^4 (-20 du)$$ Analyzing the integral, the limits of integration have now become the same. Whenever this happens, the integral is equal to 0. Therefore, $$\int_{0}^{1}(1+5x-x^5)^4 x^4 dx = 0$$ For the second integral, we do the same substitution $$\int_{0}^{1}(1+5x-x^5)^4 dx = \frac{1}{5}\int_{1}^{1}u^4 du$$ Again, the limits of integration have become equal, so the integral is equal to 0. Thus, the definite integral of the given function is: $$\int_{0}^{1}(1+5x-x^5)^4(x^4 - 1) dx = 0 - 0 = 0$$

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Most popular questions from this chapter

Assume \(\int\) is continuous on \([a, b]\). Assume also that \(\int_{a}^{b} f(x) g(x) d x=0\) for every function \(g\) that is continuous on \([\mathrm{a}, \mathrm{b}]\). Prove that \(\mathrm{f}(\mathrm{x})=0\) for all xin [a. b]

Evaluate the following integrals : (i) \(\int_{0}^{\pi / 2} \sin ^{5} x d x\) (ii) \(\int_{0}^{\frac{1}{2} \pi} \cos ^{6} x d x\)

Explain why each of the following integrals is improper. (a) \(\int_{1}^{\infty} x^{4} e^{-x^{4}} d x\) (b) \(\int_{0}^{\pi / 2} \sec x d x\) (c) \(\int_{0}^{2} \frac{x}{x^{2}-5 x+6} d x\) (d) \(\int_{-\infty}^{0} \frac{1}{x^{2}+5} d x\)

Show that \(0.78<\int_{0}^{1} \frac{d x}{\sqrt{1+x^{4}}}<0.93\)

Given that f satisfies \(|\mathrm{f}(\mathrm{u})-\mathrm{f}(\mathrm{v})| \leq|\mathrm{u}-\mathrm{v}|\) for \(\mathrm{u}\) and \(v\) in \([a, b]\) then prove that (i) \(\mathrm{f}\) is continuous in \([\mathrm{a}, \mathrm{b}]\) and (ii) \(\left|\int_{a}^{b} f(x) d x-(b-a) f(a)\right| \leq \frac{(b-a)^{2}}{2}\).
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