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Evaluate the integral between zero and two of two sin π₯ minus three π to the π₯ with respect to π₯.
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To answer this question, weβre going to be using the second part of the fundamental theorem of calculus.
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This tells us that if lowercase π is a continuous function on the closed interval between π and π and capital πΉ is any antiderivative of lowercase π.
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Then the integral between π and π of lowercase π of π₯ with respect to π₯ is equal to capital πΉ of π minus capital πΉ of π.
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Looking back at our question, the first thing we might notice is that the function lowercase π, which is our integrand, consists of two different terms.
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The first term involves the trigonometric function sine.
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And the second involves the exponential π.
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Now, we should be familiar with the fact that both trigonometric and exponential functions of this form are continuous over the entire set of real numbers.
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We have therefore fulfilled the criteria that our function π must be continuous over the closed interval between π and π, which in our case is the closed interval between zero and two.
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Now, given that we do have two terms, we might find that our working is clearer if we split these up into separate integrals.
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We would do this like so, remembering to keep the limits of integration the same across both terms.
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We can now evaluate each of these integrals separately.
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The antiderivative of two sin π₯ is negative two cos π₯.
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And the antiderivative of negative three π to the π₯ is negative three π to the π₯.
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Of course, remember, we can ignore the constant of integration in both cases since weβre working with definite integrals.
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Here, we note that weβve expressed our antiderivative in brackets, with the limits of integration being carried over to the right-hand bracket in both cases.
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Given that both of these brackets have the same limits of integration, we can simply combine them.
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Now, you might notice that we couldβve moved directly from our original integral to this step, by treating each of the terms individually.
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Instead of splitting our integral into two and then recombining, we would simply have found the antiderivative of each of our terms.
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If youβre not sure, however, thereβs no harm in writing out the method in full.
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To move forward with our question, we now substitute in the limits of our integration, which are zero and two.
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We then reach the following expression.
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With our first set of parentheses, thereβre no simplifications necessary.
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So we can just leave this.
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For the second set of parentheses, we might recall that cos of zero is equal to one.
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So negative two cos zero is equal to negative two.
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Alongside this, π to the power of zero is also one.
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So negative three π to the power of zero is negative three.
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Our second set of parentheses therefore becomes negative two minus three, which is negative five.
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But for our final answer, weβre subtracting this.
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So weβre left with a positive five.
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And weβve now reached our final answer.
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The definite integral given in the question evaluates to negative two cos of two minus three π squared plus five.