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Chapter 2: Problem 23

If \(\mathrm{f}\) is a continuous function such that \(\int_{0}^{x} f(t) d t=x e^{2 x}+\int_{0}^{x} e^{-t} f(t) d t\) for all \(\mathrm{x}\), find an explicit formula for \(\mathrm{f}(\mathrm{x})\).

Short Answer

Expert verified
Question: Find an explicit formula for the continuous function \(f(x)\) that satisfies the equation \(\int_{0}^{x} f(t) d t=x e^{2 x}+\int_{0}^{x} e^{-t} f(t) d t\). Answer: The explicit formula for \(f(x)\) is \(f(x) = \frac{e^{2x}(1+2x)}{1 - e^{-x}}\).

Step by step solution

01

Differentiate Both Sides of the Equation

To find an explicit formula for \(\mathrm{f(x)}\), we'll start by differentiating both sides of the given equation, using linearity of differentiation. By doing this, we hope to isolate \(\mathrm{f(x)}\) on one side of the equation: \(\frac{d}{dx} \left(\int_{0}^{x} f(t) d t\right) = \frac{d}{dx} \left(x e^{2 x}+\int_{0}^{x} e^{-t} f(t) d t\right)\)
02

Differentiate The Integrals

Now we will apply the Fundamental Theorem of Calculus, which states that if \(\mathrm{F}\) is an antiderivative of \(\mathrm{f}\), then the derivative of the integral of \(\mathrm{f}\) with respect to \(\mathrm{x}\) is just the function \(\mathrm{f}\) itself: \(f(x) = \frac{d}{dx} \left(x e^{2 x}+\int_{0}^{x} e^{-t} f(t) d t\right)\) Next, we differentiate the other side of the equation: \(f(x) = \frac{d}{dx}(x e^{2x}) + \frac{d}{dx} \left(\int_{0}^{x} e^{-t} f(t) d t\right)\) Autoresizing
03

Apply the Product Rule to Differentiate \(x e^{2x}\)

To differentiate the term \(x e^{2x}\), we need to use the product rule, which states that the derivative of the product of two functions is given by: \((u(x)v(x))' = u'(x)v(x) + u(x)v'(x)\) Where \(u(x) = x\) and \(v(x) = e^{2x}\). Then we compute the derivatives: \(u'(x) = \frac{d}{dx}(x) = 1\) \(v'(x) = \frac{d}{dx}(e^{2x}) = 2e^{2x}\) Now we can apply the product rule: \(\frac{d}{dx}(x e^{2x}) = u'(x)v(x) + u(x)v'(x) = 1\cdot e^{2x} + x\cdot 2e^{2x} = e^{2x}(1+2x)\)
04

Differentiate the Second Integral Using the Fundamental Theorem of Calculus

Differentiating the second integral: \(\frac{d}{dx} \left(\int_{0}^{x} e^{-t} f(t) d t\right) = e^{-x} f(x)\)
05

Re-write the Equation and Isolate \(f(x)\)

Now, we can substitute the derivatives back into our original equation: \(f(x) = e^{2x}(1+2x) + e^{-x} f(x)\) To isolate \(f(x)\), we'll subtract \(e^{-x} f(x)\) from both sides of the equation: \(f(x) - e^{-x} f(x) = e^{2x}(1+2x)\) Factor out \(f(x)\): \(f(x)(1 - e^{-x})= e^{2x}(1+2x)\) Finally, solve for \(f(x)\) by dividing both sides by \((1 - e^{-x})\): \(f(x) = \frac{e^{2x}(1+2x)}{1 - e^{-x}}\) Now we have an explicit formula for \(\mathrm{f(x)}\).

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Most popular questions from this chapter

If each case, give an example of a continuous function \(\mathrm{f}\) satisfying the conditions stated for all real \(\mathrm{x}\), or else explain why there is no such function : (a) \(\int_{0}^{x} \mathrm{f}(\mathrm{t}) \mathrm{dt}=\mathrm{e}^{x}\) (b) \(\int_{0}^{x^{2}} f(t) d t=1-2^{x^{2}}\). (c) \(\int_{0}^{x} f(t) d t f^{2}(x)-1\).

Evaluate \(\int_{0}^{\frac{1}{2} x} \frac{\cos ^{2} \theta d \theta}{a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta}=\frac{\pi}{2 a(a+b)}\) \(a, b>0 .\)

A function \(\mathrm{f}\) is defined for all real \(\mathrm{x}\) by the formula \(\mathrm{f}(\mathrm{x})=3+\int_{0}^{\mathrm{x}} \frac{1+\sin \mathrm{t}}{2+\mathrm{t}^{2}} \mathrm{dt}\). Without attempting to evaluate this integral, find a quadratic polynomial \(\mathrm{p}(\mathrm{x})=\mathrm{a}+\mathrm{bx}+\mathrm{cx}^{2}\) such that \(\mathrm{p}(0)=\mathrm{f}(0), \mathrm{p}^{\prime}(0)=\mathrm{f}^{\prime}(0)\), and \(\mathrm{p}^{\prime \prime}(0)=\) f' \((0)\).

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