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Let \(f:[0, \infty) \rightarrow \mathbb{R}\) be continuous. Find \(f(2)\) if for all \(x \geq 0\), (i) \(\int_{0}^{x} f(t) d t=x^{2}(1+x)\), (ii) \(\int_{0}^{f(x)} t^{2} d t=x^{2}(1+x)\), (iii) \(\int_{0}^{x^{2}} f(t) d t=x^{2}(1+x)\), (iv) \(\int_{0}^{x^{2}(1+x)} f(t) d x=x\).

Short Answer

Expert verified
The value of \(f(2)\) is 12.

Step by step solution

01

Differentiate the integrals using Fundamental Theorem of Calculus (FTC)

First, we need to differentiate the integrals in the given equations using the FTC. The FTC states that if \(F(x)\) is an antiderivative of \(f(x)\) on the interval \([a, b]\), then: \[ F(b) - F(a) = \int_a^b f(x) dx \] In our case, let's differentiate all the given equations with respect to \(x\) using the FTC:
02

(i) Differentiating the integral involving \(f(t)\) with respect to \(x\)

Differentiating the integral in equation (i) gives us: \( \frac{d}{dx} \left(\int_0^x f(t) dt \right) = \frac{d}{dx} \left( x^2(1+x) \right) \) By the FTC, we find: \( f(x) = 2x(1+x) + x^2 \)
03

(ii) Differentiating the integral involving \(t^2\) with respect to \(x\)

Differentiating the integral in equation (ii) gives us: \( \frac{d}{dx} \left(\int_0^{f(x)} t^2 dt \right) = \frac{d}{dx} \left( x^2(1+x) \right) \) By the FTC, we find: \( f'(x) f(x) = 2x(1+x) + x^2 \)
04

(iii) Differentiating the integral involving \(f(t)\) with respect to \(x^2\)

Differentiating the integral in equation (iii) gives us: \( \frac{d}{dx^2} \left(\int_0^{x^2} f(t) dt \right) = \frac{d}{dx^2} \left( x^2(1+x) \right) \) By the FTC, we find: \(2 f(x^2) = 2x(1+x) + x^2 \)
05

(iv) Differentiating the integral involving \(f(t)\) with respect to \(x^2(1+x)\)

Differentiating the integral in equation (iv) gives us: \( \frac{d}{dx^2(1+x)} \left(\int_0^{x^2(1+x)} f(t) dt \right) = \frac{d}{dx^2(1+x)} \left( x \right) \) By the FTC, we find: \( f(x^2(1+x)) (2x(1+x) + x^2) = 1 \)
06

Evaluate f(2)

From equation (i), we have: \( f(x) = 2x(1+x) + x^2 \) We need to evaluate \(f(2)\). So we plug in \(x = 2\) to get: \( f(2) = 2\cdot 2 (1+2) + 2^2 \) \( f(2) = 8 + 4 = 12 \) Therefore, the value of \(f(2)\) is 12.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus (FTC) plays a crucial role in bridging the gap between differentiation and integration. It's like a bridge that connects two major concepts of calculus. The FTC consists of two parts.
  • First, it establishes the relationship between the derivative and the integral of a function.
  • Second, it provides a practical technique for evaluating definite integrals.
According to the theorem, if you have a function that is continuous over a closed interval \([a, b] \), and if \(F(x)\) is an antiderivative of \(f(x)\) on that interval, then:\[F(b) - F(a) = \int_a^b f(x) \, dx\]This means that you can find the total accumulation of values of \(f(x)\) over \([a, b]\) by simply evaluating its antiderivative at the endpoints of the interval. In the given problem, differentiating the integrals using FTC simplifies the equations and allows us to express them in a form that can be solved directly. This simplification aids in finding resulting functions or values, such as \(f(2)\) in the example.
Continuous Functions
Continuous functions are essential in calculus because they ensure certain predictable behaviors without sudden jumps or breaks. Consider a continuous function \(f(x)\) over a range — it means for any two points, no breaks occur in the path drawn by \(f(x)\). This continuity is important for applying integrations and differentiations smoothly.In calculus, especially when using the FTC, continuity ensures that the antiderivative exists over the interval of interest. Thus, we can always differentiate back appropriately. For example, in the exercise, we start with a continuous function \(f:[0, \infty) \rightarrow \mathbb{R}\). This guarantees us that each integral from \(0\) to any given point behaves nicely.Continuity also means that small changes to the value of \(x\) will result in small changes in \(f(x)\), making predictions possible and mathematical manipulation convenient. This property is what allows step-by-step solving using calculus principles with confidence.
Integration Techniques
Integration techniques are numerous, each serving to solve different kinds of problems. In calculus, integration is the process of finding an antiderivative or the area under a curve. For solving the problem statement effectively, you need familiarity with various techniques and understand when to apply each. In the exercise, the primary method involves using the FTC, yet other integration techniques like substitution or partial fractions can be critical in other scenarios. Here are some typical techniques:
  • Substitution: Useful to simplify expressions before integrating.
  • Integration by parts: A way to integrate products of functions.
  • Partial fraction decomposition: Breaks down complex rational expressions.
In the exercise, all integrals were attacked by directly differentiating, rooted in the FTC. This approach revealed new forms of equations where functions like \(f(x)\) could be easily expressed in practical terms. The understanding of such techniques empowers you to tackle diverse calculus problems efficiently, and knowing when and how to apply them is key to simplifying complex integrals.

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Most popular questions from this chapter

Let \(f:[a, b] \rightarrow \mathbb{R}\) be integrable and \(\phi:[m(f), M(f)] \rightarrow \mathbb{R}\) be continuous. Show that \(\phi \circ f:[a, b] \rightarrow \mathbb{R}\) is integrable. (Hint: Given \(\epsilon>0\), find \(\delta>0\) using the uniform continuity of \(\phi\). There is a partition \(P\) of \([a, b]\) such that \(U(P, f)-L(P, f)<\delta^{2}\). Divide the sum in \(U(P, f)-L(P, f)\) into two classes depending on whether \(M_{i}(f)-m_{i}(f)\) is less than \(\delta\), or greater than or equal to \(\delta\). Use the Riemann condition for \(\phi \circ f\).)

Assuming that \(f\) is integrable on \([0,1]\), show that $$ \lim _{n \rightarrow \infty} \frac{1}{n}\left[f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\cdots+f\left(\frac{n}{n}\right)\right]=\int_{0}^{1} f(x) d x . $$

Let \(f:[a, b] \rightarrow \mathbb{R}\) be continuous and consider the function \(F:[a, b] \rightarrow \mathbb{R}\) given by \(F(x):=\int_{a}^{x} f(t) d t\) for \(x \in[a, b]\). If \(f(x) \geq 0\) for all \(x \in[a, b]\), then show that \(F\) is monotonically increasing on \([a, b]\), and if \(f\) monotonically increasing on \([a, b]\), then \(F\) is convex on \([a, b]\). (Hint: Part (i) of Proposition \(4.27\) and Part (i) of Proposition 4.31.)

Let \(c \in(a, b)\) and \(f:[a, b] \rightarrow \mathbb{R}\) be given by $$ f(x):=\left\\{\begin{array}{ll} (x-c) /(a-c) & \text { if } a \leq x \leq c, \\ (x-c) /(b-c) & \text { if } c

A bounded subset \(E\) of \(\mathbb{R}\) is said to be of (one-dimensional) content zero if the following condition holds: For every \(\epsilon>0\), there is a finite number of closed intervals whose union contains \(E\) and the sum of whose lengths is less than \(\epsilon\). Prove the following statements: (i) A subset of a set of content zero is of content zero. (ii) A finite union of sets of content zero is of content zero. (iii) If \(E\) is of content zero and \(\partial E\) denotes the boundary of \(E\), then \(E \cup \partial E\) is of content zero. (iv) A set \(E\) is of content zero if and only if the interior of \(E\) is empty and \(\partial E\) is of content zero. (v) Every finite subset of \(\mathbb{R}\) is of content zero. (vi) The infinite set \(\\{1 / n: n \in \mathbb{N}\\}\) is of content zero. (vii) The infinite set \(\mathbb{Q} \cap[0,1]\) is not of content zero.

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