Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Suppose \(\int_{1}^{x} f(t) d t=x^{2}-2 x+1 .\) Find \(f(x)\)

Short Answer

Expert verified
Question: Given the integral of a function f(t) that evaluates to x^2 - 2x + 1 over the interval [1, x], find the function f(x). Answer: f(x) = 2x - 2

Step by step solution

01

Identify the antiderivative F(x)

The antiderivative F(x) for this exercise is given by: $$ F(x) = x^2 - 2x + 1 $$
02

Take the derivative of F(x)

To find f(x), take the derivative of F(x) with respect to x: $$ f(x) = \frac{d}{dx}(x^2 - 2x + 1) $$
03

Calculate the derivative

Now, use the power rule to find the derivative of x^2 - 2x + 1: $$ f(x) = 2x - 2 $$ So, the function f(x) is: $$ f(x) = 2x - 2 $$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\sqrt{1}+x\) 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) \(0.5<\int_{0}^{1 / 2} \frac{\mathrm{dx}}{\sqrt{\left(1-\mathrm{x}^{2 \mathrm{a}}\right)}}<0.524\).

Let \(\mathrm{F}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}\). Determine a formula for computing \(\mathrm{F}(\mathrm{x})\) for all real \(\mathrm{x}\) if \(\mathrm{f}\) is defined as follows: (a) \(\mathrm{f}(\mathrm{t})=(\mathrm{t}+\mid \mathrm{t})^{2}\) (b) \(f(t)=\left\\{\begin{array}{lll}1-t^{2} & \text { if } & |t| \leq 1 \\\ 1-|t| & \text { if } & |t|>1\end{array}\right.\) (c) \(\mathrm{f}(\mathrm{t})=\mathrm{e}^{-1}\). (d) \(\mathrm{f}(\mathrm{t})=\) the maximum of 1 and \(\mathrm{t}^{2}\).

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 \(\int_{0}^{1} \frac{1}{1+x^{2}} d x>\frac{3}{4}\).

Find the greatest and least values of the function \(\mathrm{I}(\mathrm{x})=\int_{0}^{\mathrm{x}} \frac{2 \mathrm{t}+1}{\mathrm{t}^{2}-2 \mathrm{t}+2} \mathrm{dt}\) on the interval \([-1,1] .\)

Prove that (i) \(\int_{1}^{2} \frac{d x}{x^{3}+3 x+1}<\frac{1}{5}\) (ii) \(3 \sqrt{23}<\int_{2}^{5} \sqrt{3 \mathrm{x}^{3}-1} \mathrm{dx}<10 \sqrt{15}-8 \sqrt{6} / 5\) (iii) \(2<\int_{0}^{4} \frac{d x}{1+\sin ^{2} x}<4\) (iv) \(\frac{\pi}{2}<\int_{0}^{\pi / 2} \frac{\mathrm{d} \theta}{\sqrt{1-\mathrm{k}^{2} \sin ^{2} \theta}}<\frac{\pi}{2 \sqrt{1-\mathrm{k}^{2}}}\left(0<\mathrm{k}^{2}<1\right)\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free