Video: Using the Net Change Theorem

True or false? The net change theorem says that ∫_(4)^(9) √(π‘₯) dπ‘₯ = √9 βˆ’ √4.

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Video Transcript

True or false? The net change theorem says that the integral from four to nine of the square root of π‘₯ with respect to π‘₯ is equal to the square root of nine minus the square root of four.

The question wants us to use the net change theorem to evaluate the definite integral from four to nine of the square root of π‘₯ with respect to π‘₯. The net change theorem tells us that we can evaluate the definite integral from π‘Ž to 𝑏 of a rate-of-change function 𝐹 prime of π‘₯ with respect to π‘₯ as 𝐹 evaluated at 𝑏 minus 𝐹 evaluated at π‘Ž. That’s the net change of our function 𝐹.

Since we want to apply the net change theorem to the integral from four to nine of the square root of π‘₯ with respect to π‘₯, we’ll set our rate-of-change function 𝐹 prime of π‘₯ to be the square root of π‘₯. So to use the net change theorem, we need to find our function 𝐹. 𝐹 is an antiderivative of our function 𝐹 prime of π‘₯.

We can find our antiderivative of the square root of π‘₯ by using the power rule for integration. Which tells us the constants π‘Ž and 𝑛, where 𝑛 is not equal to negative one. The integral of π‘Žπ‘₯ to the 𝑛th power with respect to π‘₯ is equal to π‘Žπ‘₯ to the power of 𝑛 plus one divided by 𝑛 plus one plus a constant of integration 𝑐. We add one to the exponent and then divide by this new exponent.

So to find the antiderivative of the square root of π‘₯, we’re going to integrate the square root of π‘₯ with respect to π‘₯. We’ll do this by writing our integrand as π‘₯ to the power of a half. Applying the power rule for integration, we add one to the exponent, giving us one plus a half, and then divide by this new exponent. And then add our constant of integration 𝑐.

Finally, we can simplify one plus a half to give us three over two. And instead of dividing by three over two, we multiply by the reciprocal. This gives us two π‘₯ to the power of three over two divided by three plus the constant of integration 𝑐. And this gives us an expression for the antiderivative of the square root of π‘₯.

In fact, we don’t need to add our constant of integration 𝑐 since it will cancel out when we use it in the definite integral. So by applying the net change theorem, we have the integral from four to nine of the square root of π‘₯ with respect to π‘₯ is equal to 𝐹 evaluated at nine minus 𝐹 evaluated at four. Substituting π‘₯ is equal to nine and π‘₯ is equal to four into our antiderivative function gives us two times nine to the power of three over two divided by three minus two times four to the power of three over two divided by three.

We can calculate this to give us 18 minus 16 divided by three, which we can then calculate to be 38 divided by three. However, the question states that our answer should be the square root of nine minus the square root of four, which we can calculate to be three minus two, which is of course equal to one. And since this answer is different to the one we got by using the net change theorem, we can conclude that the statement is false.

The net change theorem does not say that the integral from four to nine of the square root of π‘₯ with respect to π‘₯ is equal to the square root of nine minus the square root of four. In fact, it says that this integral is equal to two times nine to the power of three over two over three minus two times four to the power of three over two divided by three.

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