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