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Evaluate the definite integral between one and two of π₯ squared times π₯ cubed minus three cubed with respect to π₯.
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This is not a polynomial thatβs nice to integrate using our standard rules of finding the antiderivative.
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And we certainly donβt want to distribute our parentheses and find the antiderivative for each term.
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Instead, we spot that our integral is set up in this form.
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Itβs the definite integral between some limits of π and π of some function of π of π₯ times the derivative of the inner part of that composite function.
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Here, π of π₯, the inner part of our composite function, is π₯ cubed minus three.
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And then, we have a scalar multiple of its derivative here.
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So, we use integration by substitution to evaluate the integral.
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This says that if π prime, the derivative of π, is continuous on some closed interval π to π, and π is continuous on the range of π’, which is our function π of π₯.
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Then, then the definite integral is equal to the definite integral between π of π and π of π of π of π’ with respect to π’.
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So, we let π’ be equal to the function that we defined as π of π₯.
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Itβs the function inside a function whose derivative also appears.
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So, weβre going to let π’ be equal to π₯ cubed minus three.
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Now, this is great as when we differentiate π’ with respect to π₯, we see that dπ’ by dπ₯ equals three π₯ squared.
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And in integration by substitution, we think of dπ’ and dπ₯ as differentials.
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And we can alternatively write this as dπ’ equals three π₯ squared dπ₯.
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Notice that whilst dπ’ by dπ₯ is definitely not a fraction, we do treat it a little like one in this process.
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We divide both sides by three.
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And we see that a third dπ’ equals π₯ squared dπ₯.
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So, letβs look back to our original integral.
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We now see that we can replace π₯ squared dπ₯ with a third dπ’.
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And we can replace π₯ cubed minus three with π’.
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But what do we do with our limits of one and two?
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Well, we need to replace them with π of one and π of two.
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Well, we go back to our original substitution.
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We said that π’ is equal to π₯ cubed minus three, and our lower limit is when π₯ is equal to one.
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So, thatβs when π’ is equal to one cubed minus three, which is equal to negative two.
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Our upper limit is when π₯ is equal to two.
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So, at this stage, π’ is equal to two cubed minus three, which is of course five.
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We can now replace each part of our integral with the various substitution.
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And we see that weβre going to need to work out the definite integral between negative two and five of a third π’ cubed with respect to π’.
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Remember, we can take any constant factors outside of the integral and focus on integrating π’ cubed.
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And then, we recall that we can integrate a polynomial term whose exponent is not equal to negative one by adding one to that exponent and then dividing by that value.
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So, the integral of π’ cubed is π’ to the fourth power divided by four.
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And this means our integral is equal to a third times π’ to the fourth power divided by four evaluated between negative two and five.
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Weβll substitute π’ equals five and π’ equals negative two, and find their difference.
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In this case, thatβs simply a third of five to the fourth power divided by four minus negative two to the fourth power divided by four.
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That gives us a third of 609 over four.
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And we can cancel through by three.
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And we obtain our solution to be equal to 203 over four, or 50.75.
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And so, the definite integral between one and two of π₯ squared times π₯ cubed minus three cubed with respect to π₯ is 50.75.
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In this example, we saw that we should try and choose π’ to be some factor of the integrand whose differential also occurs, albeit some scalar multiple of it.
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If thatβs not possible though, we try choosing π’ to be a more complicated part of the integrand.
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This might be the inner function in a composite function or similar.