Question Video: Identifying the Net Change Theorem | Nagwa Question Video: Identifying the Net Change Theorem | Nagwa

Question Video: Identifying the Net Change Theorem Mathematics

True or false: The net change theorem states that ∫_(0)^(2) 𝑒^(2π‘₯) dπ‘₯ = (1/2) (𝑒⁴ βˆ’ 1). [A] True [B] False

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Video Transcript

True or false, the net change theorem states that the integral from zero to two of 𝑒 to the power of two π‘₯ with respect to π‘₯ is equal to a half multiplied by 𝑒 to the fourth power minus one.

The question wants us to use the net change theorem. What does the net change theorem tell us? The net change theorem tells us that the definite integral from π‘Ž to 𝑏 of a rate-of-change function 𝑓 prime of π‘₯ with respect to π‘₯ is just equal to 𝑓 evaluated at 𝑏 minus 𝑓 evaluated at π‘Ž. It’s the net change of our function 𝑓. The question wants us to use the net change theorem to evaluate the indefinite integral from zero to two of 𝑒 to the power of two π‘₯ with respect to π‘₯. So, we’ll set 𝑓 prime of π‘₯ equal to 𝑒 to the power of two π‘₯. And since our integral is from zero to two, we’ll set the lower limit of our integral, π‘Ž, equal to zero and the upper limit of our integral, 𝑏, equal to two.

Now, all we need to apply the net change theorem is to find our function 𝑓 of π‘₯. We can see that 𝑓 of π‘₯ is an antiderivative of our derivative function 𝑓 prime of π‘₯. And we can find this antiderivative by using our integration rule. We know, for any constant 𝑛, the integral of 𝑒 raised to the power of 𝑛π‘₯ with respect to π‘₯ is equal to 𝑒 to the power of 𝑛π‘₯ divided by 𝑛 plus a constant of integration 𝑐. In our case, 𝑛 is equal to two. This means our function 𝑓 of π‘₯ is equal to 𝑒 to the power of two π‘₯ divided by two plus a constant of integration 𝑐. However, since we’re using a definite integral, the constants of integration will cancel during our working. So, we don’t actually need to worry about this.

So, by using the net change theorem, we have the definite integral from zero to two of 𝑒 to the power of two π‘₯ with respect to π‘₯ is equal to 𝑓 of two minus 𝑓 of zero. This gives us 𝑒 to the power of two times two divided by two minus 𝑒 to the power of two times zero over two. And this is the point we see if we have included our constant of integration, we would’ve added 𝑐 and subtracted 𝑐. So, the constant of integration would’ve canceled. We can simplify this to give us 𝑒 to the fourth power over two minus a half since 𝑒 to zeroth power is just equal to one. Finally, we take out a factor of a half to give a half multiplied by 𝑒 to the fourth power minus one.

Therefore, we’ve shown, by using the net change theorem, the integral from zero to two of 𝑒 to the power of two π‘₯ with respect to π‘₯ is equal to a half multiplied by 𝑒 to the fourth power minus one. Our statement was true.

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