Math Problem Statement

∫[3 to 7] t^3ln(3t)dt

Solution

To evaluate the integral

37t3ln(3t)dt,\int_{3}^{7} t^3 \ln(3t) \, dt,

we can use the method of integration by parts. Let's set it up as follows:

  1. Choose u=ln(3t)u = \ln(3t) and dv=t3dtdv = t^3 \, dt.

    • Then du=13t3dt=1tdtdu = \frac{1}{3t} \cdot 3 \, dt = \frac{1}{t} \, dt.
    • And v=t3dt=t44v = \int t^3 \, dt = \frac{t^4}{4}.

Using the integration by parts formula:

udv=uvvdu,\int u \, dv = uv - \int v \, du,

we substitute:

u=ln(3t),dv=t3dt,du=1tdt,v=t44.u = \ln(3t), \quad dv = t^3 \, dt, \quad du = \frac{1}{t} \, dt, \quad v = \frac{t^4}{4}.

Thus, the integral becomes:

37t3ln(3t)dt=[t44ln(3t)]3737t441tdt.\int_{3}^{7} t^3 \ln(3t) \, dt = \left[ \frac{t^4}{4} \ln(3t) \right]_{3}^{7} - \int_{3}^{7} \frac{t^4}{4} \cdot \frac{1}{t} \, dt.

Simplifying the integral:

=[t44ln(3t)]371437t3dt.= \left[ \frac{t^4}{4} \ln(3t) \right]_{3}^{7} - \frac{1}{4} \int_{3}^{7} t^3 \, dt.

Now, we evaluate each term separately:

  1. Evaluate [t44ln(3t)]37\left[ \frac{t^4}{4} \ln(3t) \right]_{3}^{7}:

    [ \left[ \frac{7^4}{4} \ln(21) \right] - \left[ \frac{3^4}{4} \ln(9) \right]. ]

    Calculate:

    =24014ln(21)814ln(9).= \frac{2401}{4} \ln(21) - \frac{81}{4} \ln(9).

  2. Evaluate 1437t3dt\frac{1}{4} \int_{3}^{7} t^3 \, dt:

    [ = \frac{1}{4} \left[ \frac{t^4}{4} \right]_{3}^{7} = \frac{1}{16} \left[ 7^4 - 3^4 \right]. ]

    Calculate:

    =116[240181]=116×2320=145.= \frac{1}{16} \left[ 2401 - 81 \right] = \frac{1}{16} \times 2320 = 145.

Finally, combining these results:

=24014ln(21)814ln(9)145.= \frac{2401}{4} \ln(21) - \frac{81}{4} \ln(9) - 145.

This is the evaluated form of the integral:

37t3ln(3t)dt=24014ln(21)814ln(9)145.\int_{3}^{7} t^3 \ln(3t) \, dt = \frac{2401}{4} \ln(21) - \frac{81}{4} \ln(9) - 145.

Would you like more details, or do you have any questions? Here are some related questions to consider:

  1. How do you choose uu and dvdv in integration by parts?
  2. What are other methods for integrating functions involving logarithms?
  3. How would the process change if the integral was from 0 to infinity?
  4. Can integration by parts be used more than once on a single integral?
  5. What is the physical significance of logarithmic functions in real-world applications?

Tip: Integration by parts is often useful when the integrand is a product of a polynomial and a logarithmic or exponential function.

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Math Problem Analysis

Mathematical Concepts

Integration by Parts
Definite Integral
Logarithmic Functions

Formulas

Integration by Parts Formula

Theorems

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Suitable Grade Level

Advanced College Level