Video: Applying Knowledge of Integration When the Graph of the Function Lies above or below the π‘₯ Axis

The graph of 𝑓 for βˆ’2 ≀ π‘₯ ≀ 6 consists of two semicircles, as shown in the figure. What is the value of the integral from ∫_(βˆ’ 2)^(6) 𝑓(π‘₯) dπ‘₯?

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Video Transcript

The graph of 𝑓 for negative two is less than or equal to π‘₯ is less than or equal to six consists of two semicircles, as shown in the figure. What is the value of the integral from negative two to six of 𝑓 of π‘₯ with respect to π‘₯?

When we integrate the equation of a curve between two limits, we’re finding an area. It’s the area enclosed by the graph of the function 𝑓, the π‘₯-axis, and the vertical lines at the limits of the integral. In this case, that’s the lines π‘₯ equals negative two and π‘₯ equals six. When we perform this integral then, we’re actually evaluating the shaded area. That’s the area of the semicircles.

Now, we don’t actually have to perform this integral. It isn’t the intention that we find the equation of the curve 𝑓, although we could try to find the equations of these semicircles. It also isn’t the intention that we use our knowledge of circles to actually calculate these areas. Instead, we need to think about what we know about integrals. The area we’ve been asked to find is made up of two separate regions, semicircle A and semicircle B.

We could therefore use linearity of integrals to split our integral into the sum of two integrals. The integral from negative two to two of 𝑓 of π‘₯ with respect π‘₯, which will give us the area of semicircle A. Plus the integral from two to six of 𝑓 of π‘₯ with respect to π‘₯, which would give us the area of semicircle B. However, notice that semicircle A is below the π‘₯-axis. And when we use integration to evaluate an area below the π‘₯-axis, we get a negative value.

As the two semicircles are identical β€” they both have a radius of two units β€” the value we get when we perform the integral to find the area of A will be the exact negative of the value we get when we perform the integral to find the area of B. And so, when we add these two values together, one of which is positive and one of which is negative, we will get zero.

Although the total area itself is not zero, the value of the integral is. Because the positive portion above the π‘₯-axis will be exactly cancelled out by the negative portion below the π‘₯-axis. So, by recalling the linear property of integration and the fact that using integration to find an area below the π‘₯-axis will give a negative value, we see that the integral from negative two to six of 𝑓 of π‘₯ dπ‘₯ is exactly equal to zero.

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