Video: Interpreting Integrals as Areas under Curves

The figure shows the graph of 𝑓(π‘₯) = (1/4) (π‘₯ βˆ’ 2)Β² (π‘₯ + 1). Calculate the area of the shaded region, giving your answer as a fraction.

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Video Transcript

The figure shows the graph of 𝑓 of π‘₯ equals a quarter π‘₯ minus two squared π‘₯ plus one. Calculate the area of the shaded region, giving your answer as a fraction.

To calculate the area between a curve and the π‘₯-axis, we need to integrate that function with respect to π‘₯. In this case, we’ll be integrating the function a quarter π‘₯ minus two squared π‘₯ plus one with respect to π‘₯. And that’s between the limits two and negative one, since those are the π‘₯-values which represent the upper and lower limits of the area required.

We do, however, need to be a little bit careful since integrating a function whose area lies underneath the π‘₯-axis will give a negative value. In this case so, we’re purely interested in the area above the π‘₯-axis. So, we’re fine just to integrate here. And the easiest way to integrate this function, in my opinion, is to distribute the parentheses. We could use a substitution of 𝑒 equals π‘₯ minus two, but it’s simple enough to just read these parentheses. So, let’s go ahead and do that.

We begin by distributing π‘₯ minus two all squared. By remembering that we need to write this as the product of two brackets, π‘₯ minus two and π‘₯ minus two. π‘₯ multiplied by π‘₯ is π‘₯ squared. We multiply the outer terms. We get negative two π‘₯. Then we multiply the inner terms, and we get negative two π‘₯ again. Then multiplying the last terms, we get positive four. So, we can see π‘₯ minus two all squared is equal to π‘₯ squared minus four π‘₯ plus four.

We’re then going to multiply this by π‘₯ plus one, being really careful to multiply each term in the first parenthesis by each term in the second. One way to do this is to use the grid method. π‘₯ multiplied by π‘₯ squared is π‘₯ cubed. And π‘₯ squared multiplied by one is π‘₯ squared. π‘₯ multiplied by negative four π‘₯ is negative four π‘₯ squared. And negative four π‘₯ multiplied by one is negative four π‘₯. π‘₯ multiplied by four is four π‘₯. And four multiplied by one is four. If we collect like terms, we can see that π‘₯ minus two all squared multiplied by π‘₯ plus one is π‘₯ cubed minus three π‘₯ squared plus four.

So, we need to actually integrate a quarter π‘₯ cubed minus three π‘₯ squared plus four with respect to π‘₯ between those limits two and negative one we said earlier. In fact, whenever we have a function in π‘₯ is being multiplied by some constant, it can be simpler to bring this constant outside of the integration sign. This is allowed. We will simply end up multiplying by a quarter at the end, rather than at the beginning of our calculations.

Let’s integrate π‘₯ cubed minus three π‘₯ squared plus four. To integrate π‘₯ cubed, we add one to the exponent. That gives us π‘₯ to the power of four. We then divide through by the value of that new exponent. So, π‘₯ cubed becomes π‘₯ to the power of four divided by four. Negative three π‘₯ squared becomes negative three π‘₯ cubed divided by three, which is simply negative π‘₯ cubed. And four integrates to become four π‘₯. Remember we do not need the constant of integration when integrating between two limits.

We’ll just need to evaluate between the limits two and negative one. So, we substitute two and negative one into our expression and find the difference between them. Substituting two in, we get two to the power of four divided by four minus two cubed plus four multiplied by two, which is four. And the second part is negative one to the power of four divided by four minus negative one cubed plus four multiplied by negative one. And that simplifies to negative 11 over four. And if we type all of this into our calculator, we get an answer of 27 over 16. And so, we can see that the shaded area is 27 over 16 square units.

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