### Video Transcript

Calculate the integral from zero to four of two plus the absolute value of ๐ฅ minus one with respect to ๐ฅ.

Now we can see that the integrand here involves an absolute value or modulus function. And you may be thinking, โWell, how do I integrate such a function?โ What weโre going to do is to try and write this as a piecewise function instead of a function involving the absolute value. Weโll define the entire integrand to be ๐ of ๐ฅ. And what weโre going to do is think about intervals over which ๐ฅ minus one โ thatโs the part inside the absolute value function โ is positive or negative.

The point at which this function ๐ฅ minus one changes from negative to positive is when ๐ฅ minus one is equal to zero which is when ๐ฅ is equal to one. So the two intervals we will consider are ๐ฅ is greater than one and ๐ฅ is less than one. When ๐ฅ is greater than or equal to one, ๐ฅ minus one is greater than or equal to zero; itโs positive or zero. And so, the absolute value of ๐ฅ minus one is just ๐ฅ minus one itself. However, when ๐ฅ is less than one, ๐ฅ minus one is negative. And so, the absolute value of ๐ฅ minus one will be the negative of ๐ฅ minus one.

We, therefore, defined ๐ of ๐ฅ as a piecewise function as two plus ๐ฅ minus one for ๐ฅ greater than or equal to one and two plus negative ๐ฅ minus one for ๐ฅ less than one. This simplifies to one plus ๐ฅ for ๐ฅ greater than or equal to one and three minus ๐ฅ for ๐ฅ less than one. We can also consider what the graph of ๐ of ๐ฅ will look like.

The graph of the absolute value of ๐ฅ is shown in orange and our function will be a translation of the absolute value of ๐ฅ one unit in the positive ๐ฅ-direction and two units in the positive ๐ฆ-direction. Thatโs the graph shown in blue. We can see that the area below this graph between zero and four which is whatโs defined by the integral weโve been asked to calculate will be made up of two areas: the area below the graph between zero and one and the area between one and four.

Weโll, therefore, be able to split this integral of the absolute value function into the sum of two integrals of our piecewise function: an integral with limits of zero and one and an integral with limits of one and four. We have then that this integral is equal to the integral from zero to one of three minus ๐ฅ with respect to ๐ฅ plus the integral from one to four of one plus ๐ฅ with respect to ๐ฅ. Weโve split the area up into two separate areas. And in each case, weโre integrating the piecewise function in that interval.

Now, we can perform an integration, remembering that when we integrate powers of ๐ฅ, we increase the power by one and then divide by it provided the power is not equal to negative one. Our first integral will therefore be equal to three ๐ฅ minus ๐ฅ squared over two with lower limit of zero and an upper limit of one. Our second integral will be equal to ๐ฅ plus ๐ฅ squared over two with a lower limit of one and an upper limit of four. Remember we donโt need a constant of integration as these are definite integrals.

We now substitute the limits into our first integral, giving three multiplied by one minus one squared over two minus three multiplied by zero minus zero squared over two. Thatโs just three minus a half which is equal to two and a half. For our second integral, we have four plus four squared over two minus one plus one squared over two. Thatโs four plus eight minus one minus a half which is equal to 10 and a half. Adding the two evaluated integrals together gives two and a half plus 10 and a half which is equal to 13.

So by splitting this absolute value function up into a piecewise function, weโve found that the integral from zero to four of two plus the absolute value of ๐ฅ minus one with respect to ๐ฅ is equal to 13.