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Given that the definite integral from 𝑎 to three of two 𝑥𝑑𝑥 equals five, find 𝑎.
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The first thing we need to do is integrate two 𝑥d𝑥.
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Here we have two times 𝑥 to the first power.
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To integrate, we’ll add one to the power we started with and then divide by that new power.
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Two times 𝑥 squared divided by two equals 𝑥 squared.
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And we know that 𝑥 squared from 𝑎 to three equals five.
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We take our upper bound of three, plug it in for 𝑥.
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Three squared minus the lower bound squared, 𝑎 squared, which equals five.
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Three squared equals nine.
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Minus 𝑎 squared equals five.
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From there, we subtract nine from both sides.
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Nine minus nine equals zero.
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Negative 𝑎 squared equals five minus nine, which is negative four.
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Then we’ll multiply the whole equation by negative one to get rid of those negatives on both sides of the equation.
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𝑎 squared is equal to four.
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To get rid of the square, we need to take the square root of the left side.
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And if we take the square root on the left, we need to take the square root on the right.
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The square root of 𝑎 squared equals 𝑎.
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𝑎 equals the square root of four.
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And the square root of four is plus or minus two.
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𝑎 equals plus or minus two.