Math Problem Statement

Using the method of integration by parts, evaluate the following indefinite and definite integrals.

Solution

This image contains various integrals evaluated using the method of integration by parts. I’ll go through each section and summarize the results:

1. Indefinite Integrals Using Integration by Parts

  • (a) x9ln(x)dx=110x10ln(x)1100x10+C\int x^9 \ln(x) \, dx = \frac{1}{10} x^{10} \ln(x) - \frac{1}{100} x^{10} + C
  • (b) (x+3)exdx=(x+3)exex+C\int (x + 3)e^x \, dx = (x + 3)e^x - e^x + C
  • (c) xsin(5x)dx=15xcos(5x)+125sin(5x)+C\int x \sin(5x) \, dx = \frac{1}{5} x \cos(5x) + \frac{1}{25} \sin(5x) + C

2. Definite Integrals Using Integration by Parts

  • (a) 1ex3ln(x)dx=(14x4ln(x)116x4)1e=3e4+116\int_1^e x^3 \ln(x) \, dx = \left( \frac{1}{4} x^4 \ln(x) - \frac{1}{16} x^4 \right) \Big|_1^e = \frac{3e^4 + 1}{16}
  • (b) 45(x2)exdx=((x2)exex)45=2e5e4\int_4^5 (x - 2)e^x \, dx = ((x - 2)e^x - e^x) \Big|_4^5 = 2e^5 - e^4
  • (c) 0πxcos(4x)dx=(14xsin(4x)+116cos(4x))0π=0\int_0^\pi x \cos(4x) \, dx = \left( \frac{1}{4} x \sin(4x) + \frac{1}{16} \cos(4x) \right) \Big|_0^\pi = 0

3. Multiple Applications of Integration by Parts

  • (a) (x35x2+1)sin(x)dx=(x35x2+1)cos(x)+(3x210x)sin(x)+(6x10)cos(x)6sin(x)+C\int (x^3 - 5x^2 + 1) \sin(x) \, dx = -(x^3 - 5x^2 + 1) \cos(x) + (3x^2 - 10x) \sin(x) + (6x - 10) \cos(x) - 6 \sin(x) + C
  • (b) excos(x)dx=ex2(cos(x)+sin(x))\int e^x \cos(x) \, dx = \frac{e^x}{2} (\cos(x) + \sin(x))
  • (c) 0π/2x2cos(x)dx=(x2sin(x)+2xcos(x)2sin(x))0π/2=π284\int_0^{\pi/2} x^2 \cos(x) \, dx = \left( x^2 \sin(x) + 2x \cos(x) - 2 \sin(x) \right) \Big|_0^{\pi/2} = \frac{\pi^2 - 8}{4}

These calculations leverage integration by parts, with some requiring multiple applications. If you'd like detailed explanations on any specific integral or the steps involved in using integration by parts, please let me know!

Would you like a deeper breakdown of any of these problems? Here are some questions to explore further:

  1. Why is integration by parts used for these specific integrals?
  2. How does the choice of uu and dvdv affect the process?
  3. What strategies help in handling integrals that require multiple applications of integration by parts?
  4. Can integration by parts be applied to non-polynomial functions effectively?
  5. How can definite integrals benefit from integration by parts, especially with boundary evaluations?

Tip: When choosing uu in integration by parts, functions that simplify upon differentiation, such as polynomials or logarithmic functions, are generally good candidates.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration by Parts
Definite Integrals
Indefinite Integrals

Formulas

Integration by parts: ∫u dv = uv - ∫v du
Boundary evaluation for definite integrals

Theorems

Integration by Parts Theorem

Suitable Grade Level

College Level Calculus or Advanced High School Calculus