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Determine the indefinite integral of the fifth root of π¦ squared divided by the cube root of π¦ with respect to π¦.
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In this question, weβre asked to evaluate the integral of the quotient of two functions.
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And our integrand is given in terms of π¦, and weβre integrating with respect to π¦.
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Remember, as long as our integrand and what weβre integrating with respect to share the same variable, it doesnβt matter what we call this variable.
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So, we can integrate this using all of our standard rules.
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However, as it stands, we donβt know how to integrate this function directly.
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So, weβre going to need to rewrite this in a form which we can integrate.
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And since this expression involves a lot of exponents, weβll do this by using our laws of exponents.
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The first one of our laws of exponents weβll use is the following.
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We know the πth root of π₯ is equal to π₯ to the power of one over π.
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Of course, this is only true for positive integers π.
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And in fact, we can use this to rewrite both the numerator of our integrand and the denominator.
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First, in our numerator, the value of π is five.
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So, we get π¦ squared all raised to the power of one over five.
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Then, in our denominator, we can rewrite the cube root of π¦ as π¦ to the power of one over three.
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So, weβve rewritten our integral in the following form.
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However, we still canβt evaluate this integral.
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So, weβre going to need to simplify this even further.
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We can see in our numerator we have π¦ squared all raised to the power of one over five.
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And we can rewrite this as π¦ raised to the power of some number by using our laws of exponents.
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Recall, π raised to the power of π all raised to the power of π is equal to π raised to the power of π times π.
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Weβll apply this to rewrite the numerator of our integrand.
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Our value of π is π¦, our value of π is two, and our value of π is one over five.
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This gives us a new numerator of π¦ to the power of two times one over five.
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So, now, we have the integral of π¦ to the power of two times one over five divided by one to the power of one-third with respect to π¦.
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And of course, we can simplify the exponents in our numerator.
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Two times one over five is equal to two over five.
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Now, we can see our integrand is the quotient of two exponents where π¦ is the base in both of these.
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And once again, we can simplify this by using our laws of exponents.
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We know π¦ raised to the power of π divided by π¦ raised to the power of π is equal to π¦ raised to the power of π minus π.
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In this case, our value of π is two over five and our value of π is one-third.
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So using this, we get the integral of π¦ over two over five minus one-third with respect to π¦.
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And we can simplify this.
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Evaluating our exponent, we get two over five minus one-third is equal to one over 15.
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So, after all of this manipulation, we were able to rewrite our integral as the integral of π¦ raised to the power of one over 15 with respect to π¦.
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And now, we can evaluate this integral by using the power rule for integration.
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Recall for all real constants π and π, where π is not equal to negative one, the integral of π times π¦ to the πth power with respect to π¦ is equal to π times π¦ to the power of π plus one divided by π plus one plus our constant of integration πΆ.
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So, we want to add one to our exponent of π¦ and then divide by this new exponent.
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We can see in our case the exponent of π¦ is one over 15.
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So, we need to add one to this exponent of one over 15 and then divide by this new exponent.
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This gives us π¦ to the power of one over 15 plus one all divided by one over 15 plus one.
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And of course, we need to add our constant of integration πΆ.
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And the last thing weβll do is simplify our exponent and our denominator.
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One over 15 plus one is equal to 16 over 15.
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This gives us π¦ to the power of 16 over 15 divided by 16 over 15 plus πΆ.
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And we can do one last piece of simplification.
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Instead of dividing by 16 over 15, weβll multiply by the reciprocal.
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And doing this gives us our final answer.
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Therefore, we were able to show the integral of the fifth root of π¦ squared over the cube root of π¦ with respect to π¦ is equal to 15 times π¦ to the power of 16 over 15 divided by 16 plus πΆ.