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Evaluate the double integral of 𝑥𝑦 between the limits two and zero and four and one with respect to 𝑥 and with respect to 𝑦.
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So when we’re looking to evaluate this double integral, what it means is an integral inside another integral.
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So if we rewrite our integral, what we’re gonna have is the definite integral of 𝑥𝑦 between the limits four and one and this is with respect to 𝑥.
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Then this is inside the definite integral between the limits of two and zero with respect to 𝑦.
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So the first thing we’re gonna do is integrate.
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So we’re gonna integrate 𝑥𝑦.
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But because we did this with respect to 𝑥, the 𝑦 just becomes a constant term.
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So it isn’t dealt with when we’re dealing with the integration at this stage.
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So when we integrate, we integrate our 𝑥𝑦.
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And like I said, the 𝑦 just stays constant.
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We’re gonna get 𝑥 squared over two.
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And that’s because we’ve raised the exponent of the 𝑥 term by one.
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So it becomes 𝑥 squared.
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And we divide by the new exponent, which is two.
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So let’s just remind ourselves now what we do to work out the definite integral.
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Well, if we want to work out the definite integral of 𝑓 of 𝑥 between the limits 𝑏 and 𝑎, then what we do is we integrate the 𝑓 of 𝑥, which we’re at that stage now.
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And then, we substitute in for 𝑥 the value 𝑏.
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And then, we subtract from this the value of our integral with 𝑎 substituted in instead of 𝑥.
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So when we do that, we get four squared over two cause we’ve substituted in our four cause that’s the upper bound multiplied by 𝑦 minus one squared over two multiplied by 𝑦.
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Well, for each of these terms, we’re gonna get eight 𝑦 for the first term.
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And that’s because four squared is 16.
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16 over two is eight, so eight 𝑦.
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Then, for the second term, we’re gonna get half 𝑦.
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And that’s because one squared is just one.
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So it will give us a half 𝑦.
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So now, what we have is the definite integral of 15 over two 𝑦 between the limits of two and zero.
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And that’s because if you have eight 𝑦 and you take away a half, that’s seven and a half 𝑦 or 15 over two 𝑦.
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So now we just deal with this, but this time, with respect to 𝑦.
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And it’s gonna be equal to 15 over four 𝑦 squared with limits two and zero.
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And we get that cause if you integrate 15 over two 𝑦.
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First of all, we raise the exponent of 𝑦 by one.
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So we get 𝑦 squared.
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And then, we divide by the new exponent.
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So we divide by two.
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Well, if we divide 15 over two by two, that means it’s the same as multiplying the denominator by two, which gives us four.
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So we get 15 over four 𝑦 squared.
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So now what we do is we substitute in our upper and lower bounds.
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So we get 15 over four multiplied by two squared minus 15 over four multiplied by zero squared.
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And this leads to 15 over four multiplied by four.
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And that’s because if we subtract 15 over four, multiply it by zero squared.
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Well, this is just zero.
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So what can we do now?
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Well, the fours cancel.
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And that’s because we were multiplying by four and dividing by four.
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So you can remove those.
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So therefore, we can say that if we evaluate our double integral, then the result is gonna be equal to 15.
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And as I said, we’ve now found the value of the double integral.
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But why would we use a double integral?
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Well, a double integral gives us the volume below the surface of 𝑓 of 𝑥 comma 𝑦 above a region 𝑅.
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And that region 𝑅 is a rectangular region.
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So that’s why we would use a double integral.