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Use the Maclaurin series of π to the power of π₯ to express the integral of π to the power of π₯ squared with respect to π₯ as an infinite series.
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The question wants us to use the Maclaurin series of π to the power of π₯ to represent this integral as an infinite series.
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We recall the Maclaurin series of π to the power of π₯ tells us that π to the power of π₯ is equal to the sum from π equals zero to β of π₯ to the πth power divided by π factorial.
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And this is valid for all real numbers π₯.
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We want to use this Maclaurin series for π to the power of π₯ to represent the integral of π to the power of π₯ squared with respect to π₯.
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To do this, weβre going to use the fact that if we have a Maclaurin series for a function π of π₯, which is valid on the radius of convergence π
.
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Then we can generate a new power series using this Maclaurin series, which is valid on the absolute value of π of π₯ is less than π
.
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Since the question involves π to the power of π₯ squared, weβll set our function π of π₯ to be equal to π₯ squared.
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So weβll start with our Maclaurin series for π to the power of π₯, which we know is true for all real numbers π
, which is equivalent to saying the radius of convergence is β.
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And weβre going to change all of our values of π₯ to π₯ squared.
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This gives us that π to the power of π₯ squared is equal to the sum from π equals zero to β of π₯ squared raised to the πth power divided by π factorial when the absolute value of π₯ squared is less than the radius of convergence for our Maclaurin series of π to the π₯.
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However, weβve already shown that our Maclaurin series for π to the power of π₯ has an infinite radius of convergence.
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So our power series will be true for all real numbers π₯.
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We can simplify the numerator of our summand by noticing that π₯ squared all raised to the πth power is just equal to π₯ to the power of two π.
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Next, the question wants us to find an expression for the integral of π to the power of π₯ squared with respect to π₯.
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So weβre going to integrate both sides of our equation with respect to π₯.
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This gives us the integral of π to the power of π₯ squared with respect to π₯ is equal to the integral of the sum from π equals zero to β of π₯ to the power of two π divided by π factorial with respect to π₯.
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And since weβre integrating the infinite sum of a power series, we can switch the order of our integral and our sum.
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This gives us the sum from π equals zero to β of the integral of π₯ to the power of two π divided by π factorial with respect to π₯.
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Next, weβre going to use the fact that if π is not equal to negative one and π is a constant, then to integrate π multiplied by π₯ to the πth power with respect to π₯, we add one to our exponent of π₯ and then divide by this new exponent.
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Then, we add our constant of integration π.
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Since our sum starts from π equals zero, two π is never equal to negative one.
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So our exponent of π₯ is never equal to negative one.
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And π is independent of our variable π₯.
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This means we can consider it a constant with respect to π₯.
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So we can now use our integral rule to integrate each term of our sum.
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We add one to our exponents giving us π₯ to the power of two π plus one and then we divide by two π plus one.
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This gives us the sum from π equals zero to β of π₯ to the power of two π plus one divided by π factorial multiplied by two π plus one.
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And instead of adding a constant of integration for all of our terms, weβll just add one constant of integration outside of our sum, which we will call π.
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Therefore, weβve shown that we can represent the integral of π to the power of π₯ squared with respect to π₯ as the sum from π equals zero to β of π₯ to the power of two π plus one divided by π factorial multiplied by two π plus one plus a constant of integration π.