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Evaluate the definite integral of negative four sin ๐ between the limits ๐ and ๐ on six.
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So the first thing weโre gonna do is weโre gonna take the negative four, so our constant term, outside of our integral.
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And this is because this wonโt affect the integral of sin ๐.
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So now, what we have is negative four multiplied by the definite integral of sin ๐ between the limits of ๐ and ๐ over six.
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Now just remind us what we do with definite integrals is if we want to find the definite integral of a function between ๐ and ๐, then this is equal to the integral of that function with ๐ substituted in instead of ๐ฅ minus the integral of that function with ๐ substituted in instead of ๐ฅ.
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So therefore, weโre gonna get negative four multiplied by โ and then weโve got negative cos ๐.
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And then, we still got our same limits.
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And thatโs because if you integrate sin ๐, itโs one of our known integrals, we get negative cos ๐.
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So now, what we do is we use the rule that we described earlier.
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And weโre gonna substitute in ๐ and ๐ over six for ๐.
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And when we do that, we get negative four multiplied by negative cos ๐ minus negative cos ๐ over six.
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Well, these are known values because, first of all, we know that cos ๐ is gonna be equal to negative one.
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So then, weโre gonna get negative four multiplied by one.
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And we get one because we had negative cos ๐.
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And cos ๐ was equal to negative one.
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And negative negative one is positive one.
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And then, this is minus negative root three over two.
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And thatโs because cos ๐ over six is equal to root three over two.
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And thatโs one of the standard trig values we should know.
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But if we want to think about it in degrees, because if thatโs easier for you to remember, then ๐ over six radians is equal to 30 degrees.
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And we know that cos 30 is equal to root three over two.
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Okay great.
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So now, letโs see if we can simplify this.
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Well, weโre gonna get negative four.
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And thatโs because negative four multiplied by one is negative four and then minus โ and weโve got four root three over two.
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And weโve got minus four root three over two because we had negative four multiplied by positive root three over two.
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And thatโs because we had minus and negative.
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So that became positive.
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So then, we can divide the numerator and the denominator of the second term by two.
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So therefore, we can say that the definite integral of negative four sin ๐ between the limits of ๐ and ๐ over six is equal to negative four minus two root three.