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Chapter 2: Problem 18

Determine whether the value of the integral is positive or negative. (i) \(\int_{-3}^{-1} \frac{x^{4}}{\sqrt{3-x}} d x\) (ii) \(\int_{-2}^{4} \frac{x^{3}}{|x|+1} d x\)

Short Answer

Expert verified
(i) The function \(f(x) = \frac{x^4}{\sqrt{3-x}}\) on the interval [-3, -1]. (ii) The function \(g(x) = \frac{x^3}{|x|+1}\) on the interval [-2, 4]. Answer: (i) The value of the integral for the function \(f(x) = \frac{x^4}{\sqrt{3-x}}\) on the interval [-3, -1] is positive. (ii) We cannot determine whether the integral for the function \(g(x) = \frac{x^3}{|x|+1}\) on the interval [-2, 4] is positive or negative without actually computing the integral.

Step by step solution

01

Function Sign on the Interval [-3, -1]

We start by analyzing the function \(\frac{x^4}{\sqrt{3-x}}\) on the interval \([-3, -1]\). Since \(x^4\) is always nonnegative and \(\sqrt{3-x}\) is also nonnegative in the given interval (as \(x\) will always be less than or equal to 3, ensuring that \(3-x\) is nonnegative), the integrand will always be nonnegative.
02

Conclusion for (i)

Since the function \(\frac{x^4}{\sqrt{3-x}}\) is nonnegative on the interval \([-3, -1]\), and the interval is from a lower bound of -3 to an upper bound of -1, the value of the integral \(\int_{-3}^{-1} \frac{x^{4}}{\sqrt{3-x}} d x\) will be positive. (ii)
03

Function Sign on the Interval [-2, 4]

Now, let's analyze the function \(\frac{x^3}{|x|+1}\) on the interval [-2, 4]. We need to determine the signs of \(x^3\) and \(\frac{1}{|x| + 1}\) separately since the product of their signs will ultimately determine the sign of the integral. The sign of \(x^3\) will be the same as the sign of \(x\) since the cube of a positive number will be positive and the cube of a negative number will be negative. This means that on the interval \([-2, 0)\), \(x^3\) will be negative, and on \((0, 4]\), \(x^3\) will be positive. The function \(|x|+1\) will always be positive, which means that \(\frac{1}{|x|+1}\) will also always be positive on the given interval.
04

Breaking the Integral into Two Parts

Since the sign of \(x^3\) changes from negative to positive at \(x=0\), we can split the integral at that point: $$\int_{-2}^{4} \frac{x^{3}}{|x|+1} d x = \int_{-2}^{0} \frac{x^{3}}{|x|+1} d x + \int_{0}^{4} \frac{x^{3}}{|x|+1} d x$$. In the first integral, from \(-2\) to \(0\), the value of \(x^3\) is negative, and the value of \(\frac{1}{|x| + 1}\) is positive. This means that the integrand is negative on the interval \([-2, 0)\). In the second integral, from \(0\) to \(4\), the value of \(x^3\) is positive, and \(\frac{1}{|x| + 1}\) is also positive, making the integrand positive on the interval \((0, 4]\).
05

Conclusion for (ii)

Since the integrand is negative on the interval \([-2, 0)\) and positive on the interval \((0, 4]\), we cannot directly say whether the integral \(\int_{-2}^{4} \frac{x^{3}}{|x|+1} d x\) is positive or negative without actually computing the integral.

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Most popular questions from this chapter

Evaluate the following limits: (i) \(\lim _{n \rightarrow \infty}\left(\frac{n+1}{n^{2}+1^{2}}+\frac{n+2}{n^{2}+2^{2}}+\ldots .+\frac{1}{n}\right)\) (ii) \(\lim _{n \rightarrow \infty} \frac{2^{k}+4^{k}+6^{k}+. .+(2 n)^{k}}{n^{k+1}}, k \neq-1\) (iii) \(\lim _{n \rightarrow \infty} \frac{3}{n}\left[1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+\sqrt{\frac{n}{n+9}}+\ldots . .\right.\) \(\left.\ldots+\sqrt{\frac{n}{n+3(n-1)}}\right]\) (iv) \(\lim _{n \rightarrow x} \frac{n^{2}}{\left(n^{2}+1\right)^{3 / 2}}+\frac{n^{2}}{\left(n^{2}+2^{2}\right)^{3 / 2}}+\) \(\ldots+\frac{\mathrm{n}^{2}}{\left[\mathrm{n}^{2}+(\mathrm{n}-1)^{2}\right]^{3 / 2}}\)

Starting from \(\frac{1}{1+x}-1+x-x^{2}+\ldots+x^{2 n-1}=\frac{x^{2 n}}{1+x}\) show that \(t-\frac{t^{2}}{2}+\frac{t^{2}}{3}-\ldots-\frac{t^{2 n}}{2 n} \leq \ln (1+t) \leq t-\frac{t^{2}}{2}+\frac{t^{3}}{3}-+\frac{t^{2 n+1}}{2 n+1}\) for \(\mathrm{t} \geq 0\).

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\(\sqrt{1}+x\) Prove that, if \(\mathrm{n}>1\) (i) \(0<\int_{0}^{\pi / 2} \sin ^{n+1} x d x<\int_{0}^{\pi / 2} \sin ^{n} x d x\), (ii) \(0<\int_{0}^{\pi / 4} \tan ^{n+1} x d x<\int_{0}^{\pi / 4} \tan ^{n} x d x\). (iii) \(0.5<\int_{0}^{1 / 2} \frac{\mathrm{dx}}{\sqrt{\left(1-\mathrm{x}^{2 \mathrm{a}}\right)}}<0.524\).

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