Math Problem Statement

Exercise 3: 1. (a) Show that the integral \(\int_1^{+\infty} e^{-x} \, dx\) is convergent and calculate its value. (b) Deduce that \(\int_1^{+\infty} e^{-x^2} \, dx\) is convergent. 2. Show that the following integrals are convergent and calculate their values: (a) \(I = \int_0^1 \ln x \, dx\) (b) \(J = \int_0^1 \frac{x^2}{\sqrt{1-x^2}} \, dx\), where \(x = \sin t\), \(t \in [0, \frac{\pi}{2}]\).

Solution

Let’s break down and solve each part of this exercise step by step.


1. (a) Show that 1+exdx\int_1^{+\infty} e^{-x} \, dx is convergent and calculate its value.

Solution:

The integral is: 1+exdx.\int_1^{+\infty} e^{-x} \, dx.

The general formula for the integral of exe^{-x} is: exdx=ex+C.\int e^{-x} \, dx = -e^{-x} + C.

Evaluate the definite integral: 1+exdx=[ex]1+.\int_1^{+\infty} e^{-x} \, dx = \left[-e^{-x}\right]_1^{+\infty}.

  1. As x+x \to +\infty, ex0e^{-x} \to 0.
  2. At x=1x = 1, ex=e1-e^{-x} = -e^{-1}.

Thus: 1+exdx=0(e1)=e1.\int_1^{+\infty} e^{-x} \, dx = 0 - (-e^{-1}) = e^{-1}.

The value is: 1+exdx=1e.\int_1^{+\infty} e^{-x} \, dx = \frac{1}{e}.


1. (b) Deduce that 1+ex2dx\int_1^{+\infty} e^{-x^2} \, dx is convergent.

Solution:

The function ex2e^{-x^2} decreases faster than exe^{-x} as x+x \to +\infty. Since 1+exdx\int_1^{+\infty} e^{-x} \, dx converges, by comparison, 1+ex2dx\int_1^{+\infty} e^{-x^2} \, dx also converges by the comparison test.


2. (a) Show that 01lnxdx\int_0^1 \ln x \, dx is convergent and calculate its value.

Solution:

The integral is: I=01lnxdx.I = \int_0^1 \ln x \, dx.

Using integration by parts: lnxdx=xlnxx1xdx=xlnxx+C.\int \ln x \, dx = x \ln x - \int x \, \frac{1}{x} \, dx = x \ln x - x + C.

Evaluate the definite integral: I=[xlnxx]01.I = \left[x \ln x - x\right]_0^1.

  1. At x=1x = 1: xlnxx=1ln11=1.x \ln x - x = 1 \ln 1 - 1 = -1.

  2. At x=0x = 0: limx0+(xlnx)=limx0+lnx1/x=0(using L’Hopital’s Rule).\lim_{x \to 0^+} (x \ln x) = \lim_{x \to 0^+} \frac{\ln x}{1/x} = 0 \quad \text{(using L'Hopital's Rule)}.

Thus: I=10=1.I = -1 - 0 = -1.

The value is: I=1.I = -1.


2. (b) Show that 01x21x2dx\int_0^1 \frac{x^2}{\sqrt{1-x^2}} \, dx is convergent and calculate its value.

Solution:

Substitute x=sintx = \sin t, so dx=costdtdx = \cos t \, dt, and 1x2=cost\sqrt{1-x^2} = \cos t. The limits transform as:

  • When x=0x = 0, t=0t = 0.
  • When x=1x = 1, t=π2t = \frac{\pi}{2}.

The integral becomes: J=01x21x2dx=0π/2sin2tcostcostdt.J = \int_0^1 \frac{x^2}{\sqrt{1-x^2}} \, dx = \int_0^{\pi/2} \frac{\sin^2 t}{\cos t} \cos t \, dt.

Simplify: J=0π/2sin2tdt.J = \int_0^{\pi/2} \sin^2 t \, dt.

Use the identity sin2t=1cos(2t)2\sin^2 t = \frac{1 - \cos(2t)}{2}: J=0π/21cos(2t)2dt=120π/21dt120π/2cos(2t)dt.J = \int_0^{\pi/2} \frac{1 - \cos(2t)}{2} \, dt = \frac{1}{2} \int_0^{\pi/2} 1 \, dt - \frac{1}{2} \int_0^{\pi/2} \cos(2t) \, dt.

  1. Compute 0π/21dt\int_0^{\pi/2} 1 \, dt: 0π/21dt=π2.\int_0^{\pi/2} 1 \, dt = \frac{\pi}{2}.

  2. Compute 0π/2cos(2t)dt\int_0^{\pi/2} \cos(2t) \, dt: cos(2t)dt=sin(2t)2.\int \cos(2t) \, dt = \frac{\sin(2t)}{2}. Evaluate from 00 to π2\frac{\pi}{2}: 0π/2cos(2t)dt=sin(π)2sin(0)2=0.\int_0^{\pi/2} \cos(2t) \, dt = \frac{\sin(\pi)}{2} - \frac{\sin(0)}{2} = 0.

Thus: J=12(π20)=π4.J = \frac{1}{2} \left(\frac{\pi}{2} - 0\right) = \frac{\pi}{4}.

The value is: J=π4.J = \frac{\pi}{4}.


Summary of Results:

  1. (a) 1+exdx=1e\int_1^{+\infty} e^{-x} \, dx = \frac{1}{e}.
  2. (b) 1+ex2dx\int_1^{+\infty} e^{-x^2} \, dx is convergent.
  3. (a) 01lnxdx=1\int_0^1 \ln x \, dx = -1.
  4. (b) 01x21x2dx=π4\int_0^1 \frac{x^2}{\sqrt{1-x^2}} \, dx = \frac{\pi}{4}.

Would you like further details on any step or explanation? Here are some related questions to consider:

  1. How does the comparison test for convergence work?
  2. What are the conditions for applying integration by parts?
  3. Why does the substitution x=sintx = \sin t simplify integrals involving 1x2\sqrt{1-x^2}?
  4. Can ex2e^{-x^2} be integrated analytically over [1,+)[1, +\infty)?
  5. How does L'Hopital's Rule apply to limits like limx0+xlnx\lim_{x \to 0^+} x \ln x?

Tip: Always check for the behavior of the integrand at the boundaries to assess convergence!

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Math Problem Analysis

Mathematical Concepts

Convergence of improper integrals
Exponential functions
Natural logarithm integration
Trigonometric substitution in integrals

Formulas

\(\int e^{-x} \, dx = -e^{-x} + C\)
Integration by parts: \(\int u \, dv = uv - \int v \, du\)
Trigonometric identities: \(\sin^2 t = \frac{1 - \cos(2t)}{2}\)

Theorems

Comparison Test for convergence
L'Hopital's Rule for evaluating limits

Suitable Grade Level

Undergraduate calculus