WEBVTT
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Determine the integral from one to π of the natural logarithm of π₯ divided by π₯ with respect to π₯.
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Weβre given a definite integral which weβre asked to evaluate.
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We can see our integrand is the natural logarithm of π₯ over π₯.
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And this is not easy to integrate.
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We donβt know an antiderivative for this expression.
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And this is the quotient of two functions.
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And we donβt know how to integrate this directly.
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So, weβre going to need to find some other way of integrating this expression.
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We might be tempted to use integration by parts.
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However, in this case, weβll see that integration by substitution is easier.
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And the reason integration by substitution will be easier is to take a look at our numerator, the natural logarithm of π₯.
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If we were to try integrating this by using the substitution π’ is equal to the natural logarithm of π₯, then by differentiating both sides of this equation with respect to π₯, we would see that dπ’ by dπ₯ is one over π₯.
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And this is a useful result.
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Since one over π₯ appears in our integrand, it will help simplify our integral.
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So, weβre ready to start using our substitution.
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First, although we know dπ’ by dπ₯ is not a fraction, when weβre using integration by substitution, it can help to think of it a little bit like a fraction.
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This gives us the equivalent statement in terms of differentials dπ’ is equal to one over π₯ dπ₯.
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Next, remember, weβre evaluating a definite integral by using integration by substitution.
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So, we need to rewrite our limits of integration.
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We do this by substituting our limits of integration into our expression for π’.
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To find our new upper limit of integration, we substitute π₯ is equal to π into this expression.
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This gives us π’ is equal to the natural logarithm of π, which we know is equal to one.
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And we do the same to find our new lower limit of integration.
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We substitute in π₯ is equal to one, giving us π’ is equal to the natural logarithm of one, which we know is equal to zero.
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So, weβre now ready to rewrite our integral by using our substitution.
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First, we showed the new limits of integration would be lower limit zero, upper limit one.
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Next, weβll replace the natural logarithm of π₯ with π’.
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And then finally, we know that one over π₯ dπ₯ is equal to dπ’.
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So, weβve rewritten our integral as the integral from zero to one of π’ with respect to π’.
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And we can just evaluate this integral by using the power rule for integration.
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The integral of π’ with respect to π’ is π’ squared over two.
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Then, all we do is evaluate this at the limits of integration, giving us one squared over two minus zero squared over two, which we can calculate is just equal to one-half.
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Therefore, by using the substitution π’ is equal to the natural logarithm of π₯, we were able to show the integral from one to π of the natural logarithm of π₯ over π₯ with respect to π₯ is equal to one-half.