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Chapter 2: Problem 2

Show that for any number \(\mathrm{a}>1\) \(\int_{1}^{a} \ln x d x+\int_{0}^{\ln a} \mathrm{e}^{y} d y=a \ln a\)

Short Answer

Expert verified
Question: Show that for any number \(a > 1\), \(\int_{1}^{a} \ln x dx + \int_{0}^{\ln a} e^y dy = a \ln a\).

Step by step solution

01

Evaluate the first integral \(\int_{1}^{a} \ln x dx\)

To evaluate this integral, we'll use integration by parts. Integration by parts states that \(\int u dv = uv - \int v du\). We'll choose \(u = \ln x\) and \(dv = dx\). So, \(du = \frac{1}{x}dx\) and \(v = x\). Now, substituting these into the integration by parts formula, we get: \(\int \ln x dx = x \ln x - \int x \cdot \frac{1}{x} dx\) Now we can evaluate this integral: \(\int \ln x dx = x \ln x - \int dx = x \ln x - x + C\) We can now evaluate the definite integral: \(\int_{1}^{a} \ln x dx = \left[a \ln a - a\right] - \left[\ln 1 - 1\right] = a\ln a - a + 1\)
02

Evaluate the second integral \(\int_{0}^{\ln a} e^y dy\)

To evaluate the integral \(\int e^y dy\), we know that the integral of \(e^y\) with respect to \(y\) is just \(e^y\): \(\int e^y dy = e^y + C\) Now, we can evaluate the definite integral: \(\int_{0}^{\ln a} e^y dy = \left[e^{\ln a}\right] - \left[e^{0}\right] = a - 1\)
03

Sum the two integrals and check if it equals \(a \ln a\)

Now, let's sum the two integrals and check if it equals \(a \ln a\): \((a\ln a - a + 1) + (a - 1) = a\ln a - a + 1 + a - 1 = a\ln a\) Thus, we have shown that for any number \(a > 1\), \(\int_{1}^{a} \ln x dx + \int_{0}^{\ln a} e^y dy = a \ln a\).

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Most popular questions from this chapter

If a is positive and \(\mathrm{I}=\int_{-1}^{1} \frac{\mathrm{dx}}{\sqrt{1-2 \mathrm{ax}+\mathrm{a}^{2}}}\) then show that \(\mathrm{I}=2 \mathrm{ifa}<1\) and \(\mathrm{I}=\frac{2}{\mathrm{a}}\) if \(\mathrm{a}>1\).

(a) Make a conjecture about the value of the limit \(\lim _{k \rightarrow 0} \int_{1}^{b} t^{k-1} d t(b>0)\) (b) Check your conjecture by evaluating the integral and finding the limit. [Hint: Interpret the limit as the definition of the derivative of an exponential function]

One of the numbers \(\pi, \pi / 2,35 \pi / 128,1-\pi\) is the correct value of the integral \(\int_{0}^{\pi} \sin ^{8} x d x\). Use the graph of \(\mathrm{y}=\sin ^{8} \mathrm{x}\) and a logical process of elimination to find the correct value.

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' \((0)\).

If \(\mathrm{I}=\int_{0}^{1} \frac{\mathrm{dx}}{1+\mathrm{x}^{3 / 2}}\), prove that, \(\ell \mathrm{n} 2<\mathrm{I}<\frac{\pi}{4}\).
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