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Chapter 2: Problem 10

Let \(\mathrm{f}(\mathrm{t})\) be a function that is continuous and satisfies \(\mathrm{f}(\mathrm{t}) \geq 0\) on the interval \(\left[0, \frac{\pi}{2}\right)\). Suppose it is known that for any number \(x\) between 0 and \(\frac{\pi}{2}\), the region under the graph of \(\mathrm{f}\) on \([0, \mathrm{x}]\) has area \(\mathrm{A}(\mathrm{x})=\tan \mathrm{x}\) (a) Explain why \(\int_{0}^{x} f(t) d t=\tan x\) for \(0 \leq x<\frac{\pi}{2}\) (b) Differentiate both sides of the equation in part (a) and deduce the formula of \(f\).

Short Answer

Expert verified
Answer: f(t) = sec^2(t)

Step by step solution

01

Interpret the problem

We are given the area under the curve of function f(t) from 0 to x. Since A(x) = tan(x), we can write the definite integral of f(t) from 0 to x as follows: 鈭�(0 to x) f(t) dt = tan(x)
02

Differentiate both sides

Differentiate both sides of the equation with respect to x: d/dx[鈭�(0 to x) f(t) dt] = d/dx[tan(x)]
03

Apply the Fundamental Theorem of Calculus

By applying the Fundamental Theorem of Calculus, we can rewrite the left side of the equation as follows: f(x) = d/dx[tan(x)]
04

Differentiate tan(x)

Differentiate tan(x) with respect to x: f(x) = sec^2(x)
05

Write the final answer

From these steps, we deduced the formula for the function f(t) as follows: f(t) = sec^2(t)

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Most popular questions from this chapter

It is known that \(\int_{a}^{b} f(x) d x>\int_{a}^{b} g(x) d x\). Does it follow that \(\mathrm{f}(\mathrm{x}) \geq \mathrm{g}(\mathrm{x}) \forall \mathrm{x} \in[\mathrm{a}, \mathrm{b}] ?\) Give examples.

Using Schwartz-Bunyakovsky inequality with \(\mathrm{f}^{2}(\mathrm{x})=\frac{1}{1+\mathrm{x}^{2}}, \mathrm{~g}^{2}(\mathrm{x})=1+\mathrm{x}^{2}\), show that \(\int_{0}^{1} \frac{1}{1+x^{2}} d x>\frac{3}{4}\).

Prove that \(\int_{0}^{\pi / 2} \cos ^{\mathrm{m}} \mathrm{x} \sin ^{\mathrm{m}} \mathrm{xd} \mathrm{x}=2^{-\mathrm{m}} \int_{0}^{\pi / 2} \cos ^{\mathrm{m}} \mathrm{xdx} .\)

If \(\alpha\) and \(\phi\) are positive acute angles then prove that \(\phi<\int_{0}^{p} \frac{\mathrm{dx}}{\sqrt{\left(1-\sin ^{2} \alpha \sin ^{2} \mathrm{x}\right)}}<\frac{\varphi}{\sqrt{\left(1-\sin ^{2} \alpha \sin ^{2} \varphi\right)}} .\) If \(\alpha=\phi=1 / 6 \pi\), then prove that the integral lies between \(0.523\) and \(0.541\).

Prove that \(\lim _{\lambda \rightarrow \infty} \int_{0}^{\infty} \frac{1}{1+\lambda x^{4}} d x=0\).
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