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Find the definite integral of six sin ππ§ over four between the limits negative seven and negative five.
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So the first thing we do is we can take our six outside of our integration.
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And we do that because thatβs a constant term.
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And it wouldnβt affect our integration.
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So now, what we have is six multiplied by the definite integral of sin ππ§ over four between the limits of negative seven and negative five.
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So now, to remind us what we do when weβre looking to find a definite integral.
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So if we want to find a definite integral between the limits π and π of a function, then what we do is we integrate that function, substituting π for π₯ and π for π₯.
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And then, we subtract the value with π substituted in from the value with π substituted in.
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So now, what weβre gonna need to do is integrate sin ππ§ over four.
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And to complete this, what Iβm gonna do is use substitution.
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So Iβm gonna say that π’ is equal to ππ§ over four.
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So now, if we differentiate this with respect to π§, what we get is dπ’ dπ§ is equal to π over four.
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And we get that because if weβre differentiating ππ§ over four, itβs the same as π over four multiplied by the derivative of π§, where the derivative of π§ is just gonna be one.
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So therefore, we get our π over four.
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And in integration by parts, we think of dπ’ and dπ§ as differentials.
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So then, we can alternatively write this as dπ’ is equal to π over four dπ§.
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And we can see that although dπ’ over dπ§ is definitely not a fraction, we do treat it a little bit like one in this process.
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So then, what we do is we divide through by π over four, which is the same as multiplying by four and dividing by π.
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So we get four over π. dπ’ is equal to dπ§.
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So great, we now have dπ§ in terms of dπ’.
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So we can substitute this back into our integration.
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And what weβre gonna have is six multiplied by the definite integral of four over π sin π’ dπ’ between the limits negative seven and negative five.
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So once again, we can make things easier by taking out our coefficient.
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So we got a constant term.
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So what we have is 24 over π.
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And we get that because six multiplied by four over π is 24 over π multiplied by the definite integral of sin π’ dπ’ between the limits negative seven and negative five.
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So if we integrate sin π’, we get negative cos π’.
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This is one of our standard integrations because we know if we integrate sin π₯, we get negative cos π₯.
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And at this stage, we could also use the definition of our substitution to change our limits.
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However, Iβm going to replace π’ with ππ§ over four after integration, which has the same effect.
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So now, what we need to do is substitute back in our π’.
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So weβve got negative cos ππ§ over four instead of negative cos π’.
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So this is gonna give us 24 over π multiplied by.
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Then, weβve got negative cos of negative five π over four minus negative cos of negative seven π over four.
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And weβve got that cause weβve substituted in our negative seven and negative five for π§.
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So now, before we calculate this, we need to make sure that our calculator is in the correct setting.
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And we need to make sure itβs in radians.
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It should have a little r or a little rad in the corner of the display.
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And thatβs because weβre dealing with the angles that are in terms of π.
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So we know that weβre dealing with radians.
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So itβs gonna give us 24 over π multiplied by root two over two minus negative root two over two.
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So this brings us to our final answer, which is 24 root two over π.
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And we got that because if we had root two over two minus negative root two over two.
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Well, if we subtract a negative, itβs the same as adding.
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So we get two root two over two.
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Well, if we have two root two over two, the twos will cancel.
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So we just get root two.
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So then, we get 24 over π multiplied by root two, which gives us 24 root two over π.