Question Video: Evaluating the Definite Integral of an Absolute Value Function | Nagwa Question Video: Evaluating the Definite Integral of an Absolute Value Function | Nagwa

Question Video: Evaluating the Definite Integral of an Absolute Value Function Mathematics

If 𝑓(π‘₯) = 6(π‘₯), determine ∫^(6)_(βˆ’6) 𝑓(π‘₯) dπ‘₯.

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Video Transcript

If 𝑓 of π‘₯ is equal to six times the absolute value of π‘₯, determine the integral from negative six to six of 𝑓 of π‘₯ with respect to π‘₯.

In this question, we’re asked to evaluate the definite integral of six times the absolute value of π‘₯. And we know a few different ways for evaluating a definite integral. For example, we could use the fundamental theorem of calculus. However, in this instance, it is actually easier to just determine the area under our curve. There’s a few different ways we can approach this. The first thing we’re going to want to do is sketch a graph 𝑦 is equal to six times the absolute value of π‘₯. And there’s two ways we could do that. For example, we could use a piecewise definition for the absolute value of π‘₯ and then use this to sketch the graph of 𝑦 is equal to six times the absolute value of π‘₯. And this would work.

However, we can also sketch this by realizing this is a vertical stretch of factor six of the curve 𝑦 is equal to the absolute value of π‘₯. In fact, we can also notice this is a horizontal stretch of factor one-sixth on the curve 𝑦 is equal to the absolute value of π‘₯. Either method would work. Now let’s use this to determine the definite integral from negative six to six of 𝑓 of π‘₯. To do this, we’ll start by adding π‘₯ is equal to six and π‘₯ is equal to negative six onto our sketch. And the next thing we need to do is recall that the definite integral of a continuous function gives us the signed area under the curve between these two limits.

Since we know that six times the absolute value of π‘₯ is a continuous function and we can see from our sketch this lies entirely above the π‘₯-axis, this would just be the area of our two triangles. So we want to find the area of these two triangles. This means we’re going to want to find their height. And since these points lie on the curve 𝑦 is equal to the absolute value of π‘₯, we can just substitute these values of π‘₯ in to find the height. Substituting in π‘₯ is equal to six and π‘₯ is equal to negative six, we see that we get our 𝑦-coordinates are both equal to 36. And now we can just find the area of our two triangles. Remember, the area of a triangle is one-half the base times the height.

We’ll start with the triangle from π‘₯ is equal to zero to π‘₯ is equal to six. We can find this area either by using the integral from zero to six of six times the absolute value of π‘₯ with respect to π‘₯ or by using one-half the base times the height. This gives us one-half times six times 36. And if we evaluate this, we get 108. We can do exactly the same thing for the triangle from π‘₯ is equal to negative six to π‘₯ is equal to zero. And in fact, its height and base are the same size, so we also get the area of this triangle is 108. Now, all we need to do is combine the area of these two triangles together.

Therefore, we’ve shown the integral from negative six to six of six times the absolute of π‘₯ with respect to π‘₯ is equal to the sum of the area of these two triangles, which is 108 plus 108, which we can calculate is equal to 216. In this question, we were asked to find the definite integral of a function. And instead of using the fundamental theorem of calculus, we saw that we can instead do this graphically. This meant we didn’t need to find any antiderivatives, and instead all we needed to use was our rules for triangles.

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