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Chapter 7: Problem 35

Evaluate the following definite integrals. $$\int_{1}^{e^{2}} x^{2} \ln x d x$$

Short Answer

Expert verified
The definite integral of the function $x^2\ln x$ from $1$ to $e^2$ is $\frac{5e^6}{9} + \frac{1}{9}$.

Step by step solution

01

Integration by parts formula

First, remember the integration by parts formula: $$\int u dv=uv-\int v du$$ We need to choose \(u\) and \(dv\) and differentiate or integrate them to find \(du\) and \(v\) respectively.
02

Choose u and dv

In this situation, let's choose \(u\) and \(dv\) as follows: $$u=\ln{x}$$ and $$dv=x^{2} dx$$ Find \(du\) and \(v\): $$du=\frac{1}{x} dx$$ and integrate \(dv\) to find $$v=\frac{x^{3}}{3}$$
03

Apply integration by parts

Now, apply the integration by parts formula: $$\int_{1}^{e^{2}} x^{2} \ln x d x = \left[\frac{x^{3}}{3}\ln x\right]_{1}^{e^{2}} - \int_{1}^{e^2} \frac{x^{3}}{3} \frac{1}{x} dx$$
04

Simplify the integral

Simplify the integral expression: $$\int_{1}^{e^{2}} x^{2} \ln x d x = \left[\frac{x^{3}}{3}\ln x\right]_{1}^{e^{2}} - \int_{1}^{e^2} \frac{x^{2}}{3} dx$$
05

Evaluate the integral

Evaluate the remaining integral and apply the limits of integration: $$\int_{1}^{e^{2}} x^{2} \ln x d x = \left[\frac{x^{3}}{3}\ln x\right]_{1}^{e^{2}} - \left[\frac{x^{3}}{9}\right]_{1}^{e^2}$$ Now, plug in the upper and lower limits of integration: $$\int_{1}^{e^{2}} x^{2} \ln x d x = \left(\frac{(e^{2})^{3}}{3}\ln{(e^{2})} - \frac{(e^{2})^{3}}{9}\right) - \left(\frac{1^{3}}{3}\ln{1} - \frac{1^{3}}{9}\right)$$
06

Simplify and find the final answer

Simplify the expression and get the final answer: $$\int_{1}^{e^{2}} x^{2} \ln x d x = \left(\frac{e^{6}}{3}(2) - \frac{e^{6}}{9}\right) - \left(0 - \frac{1}{9}\right)$$ Combining terms gives: $$\int_{1}^{e^{2}} x^{2} \ln x d x = \frac{2e^{6}}{3} - \frac{e^{6}}{9} + \frac{1}{9}$$ Hence, $$\int_{1}^{e^{2}} x^{2} \ln x d x = \frac{5e^{6}}{9} + \frac{1}{9}$$

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Most popular questions from this chapter

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ Verify relation \(A\) by differentiating \(x=2 \tan ^{-1} u\). Verify relations \(B\) and \(C\) using a right-triangle diagram and the double-angle formulas $$\sin x=2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right) \text { and } \cos x=2 \cos ^{2}\left(\frac{x}{2}\right)-1$$

Graph the integrands and then evaluate and compare the values of \(\int_{0}^{\infty} x e^{-x^{2}} d x\) and \(\int_{0}^{\infty} x^{2} e^{-x^{2}} d x.\)

Recall that the substitution \(x=a \sec \theta\) implies that \(x \geq a\) (in which case \(0 \leq \theta<\pi / 2\) and \(\tan \theta \geq 0\) ) or \(x \leq-a\) (in which case \(\pi / 2<\theta \leq \pi\) and \(\tan \theta \leq 0\) ). Evaluate for \(\int \frac{\sqrt{x^{2}-1}}{x^{3}} d x,\) for \(x>1\) and for \(x<-1\)

Graph the function \(f(x)=\frac{\sqrt{x^{2}-9}}{x}\) and consider the region bounded by the curve and the \(x\) -axis on \([-6,-3] .\) Then evaluate \(\int_{-6}^{-3} \frac{\sqrt{x^{2}-9}}{x} d x .\) Be sure the result is consistent with the graph.

Three cars, \(A, B,\) and \(C,\) start from rest and accelerate along a line according to the following velocity functions: $$v_{A}(t)=\frac{88 t}{t+1}, \quad v_{B}(t)=\frac{88 t^{2}}{(t+1)^{2}}, \quad \text { and } \quad v_{C}(t)=\frac{88 t^{2}}{t^{2}+1}$$ a. Which car has traveled farthest on the interval \(0 \leq t \leq 1 ?\) b. Which car has traveled farthest on the interval \(0 \leq t \leq 5 ?\) c. Find the position functions for the three cars assuming that all cars start at the origin. d. Which car ultimately gains the lead and remains in front?
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