91Ó°ÊÓ

Integration by parts is a useful technique for evaluating integrals where the integrand is a product of two functions. The general formula is given by: \ \( \ \int u \, dv = uv - \, \int v \, du \ \). \ \ Here, we pick one function to differentiate (u) and another to integrate (dv). \ We then differentiate u to find du and integrate dv to find v. \ Finally, we substitute these into our formula. This method breaks down a complex integral into simpler parts. \ For example, in our exercise: \ \( \ \int_{1}^{e} \, \ln x \, dx \ \), \ we chose: \ \ \( u = \ln x \) \ and \ \( dv = dx \), \ which gives us \ \( du = \frac{1}{x} dx \) \ and \ \( v = x \). Substituting these into the formula results in: \ \( \ x \, \ln x - \, \int x \, \frac{1}{x} \, dx \).\ This simplifies further to: \( \ x \, \ln x - \, x \). This approach simplifies our work significantly.

Definite integrals calculate the area under a curve within specified limits. They have upper and lower bounds, called limits of integration. \ For example, consider the definite integral: \ \(\ \int_{a}^{b} f(x) dx \) \ where \ a \ and \ b \ are the limits. \ The process of solving definite integrals involves: \ \

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• Finding the indefinite integral (antiderivative).
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• Evaluating this antiderivative at the upper limit.
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• Evaluating it at the lower limit.
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• Subtracting the lower limit value from the upper limit value. \
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\In our exercise, the definite integral was \ \( \ \int_{1}^{e} \, \ln x \, dx \) \. \ We found: \ \ \( x \, \ln x - \ x \), \ and then evaluated it at \ \( e \) and \ \( 1 \). \ Finally, we subtracted these values to get the area under \ \( \ln x \) from \ \( 1 \) to \ \( e \).

The natural logarithm, denoted as \(\bf \ln x \bf \), is the logarithm to the base \(\bf e \bf \) (approximately 2.71828). \ It has special properties that make it useful in calculus and integration.\ \ Some key properties are: \

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• \ \(\ln 1 = 0 \)
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• \ \(\ \ln e = 1 \)
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• The derivative of \ \(\ln x \) is \ \(\ \frac{1}{x} \)
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• \ The integral of \ \( \ ln x \, dx \) requires integration by parts.
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\ Using these properties helps simplify our work with logarithmic functions during integration. \ In the given problem, we used \ \( \ u = \ln x \) \ and found the derivative ∫ du as ∫\frac{1}{x}dx. This step-by-step use of properties and rules makes it easier to solve.