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If the rate of change of a quantity over the closed interval from one to two is given by π prime of π₯ equals π₯ squared times π to the power of two π₯ cubed, what is the net change of the quantity over the closed interval one to two?
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To answer this question, weβre going to need to quote the net change theorem.
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This says that the integral of a rate of change is equal to the net change.
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That is the definite integral between π and π of capital πΉ prime of π₯ with respect to π₯ is equal to capital πΉ of π minus capital πΉ of π.
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Now, weβre told that the rate of change of our quantity over the closed interval from one to two is given by π prime of π₯, which is equal to π₯ squared times π to the power of two π₯ cubed.
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And it doesnβt matter that weβre dealing with a lowercase π for our function.
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As long as we ensure that π of π₯ is equal to the antiderivative of π prime of π₯.
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And so, to find the net change of our quantity over the closed interval from one to two, weβre going to integrate the rate of change.
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So, itβs the definite integral between one and two.
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Those are the bounds of our interval of π₯ squared times π to the power of two π₯ cubed with respect to π₯.
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Now, this might look like a really complicated integral.
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But if we spot that π₯ squared is a scalar multiple of the derivative of two π₯ cubed, which is part of a composite function, then that tells us we can use integration by substitution.
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Letβs let π’ be equal to two π₯ cubed.
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Now, we could just let π’ be equal to π₯ cubed, where weβll get the same answer either way.
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And we choose π’ to be equal to two π₯ cubed.
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Because when we differentiate π’ with respect to π₯ here, we get six π₯ squared.
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Remember, we multiply the entire term by the exponent and then reduce the exponent by one.
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And then, whilst dπ’ by dπ₯ isnβt a fraction, we do treat it a little like one.
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And we rearrange this expression to get a sixth dπ’ equals π₯ squared dπ₯.
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And now, we replace π₯ squared dπ₯ with a sixth dπ’.
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And we replace two π₯ cubed with π’.
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But what about the limits?
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Well, we go back to our earlier substitution, we know π’ was equal to two π₯ cubed.
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And we know that when π₯ is equal to two, π’, therefore, must be equal to two times two cubed.
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And that gives us an upper limit then of 16.
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Then, our lower limit is when π₯ is equal to one.
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So, we get π’ is equal to two times one cubed, which is equal to two.
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And we now have a really nice expression to integrate.
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The integral of π to the power of π’ is π to the power of π’.
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And the integral, therefore, of a sixth π to the power of π’ is a sixth π to the power of π’.
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All thatβs left is to substitute in π’ equals 16 and π’ equals two.
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And when we do, we find that the net change of our quantity over the closed interval from one to two must be a sixth times π to the power of 16 minus a sixth times π squared.