Video: Evaluating a Definite Integral Using the Property of Reversing the Limits of Integration

If ∫_(βˆ’7)^(8) (𝑔(π‘₯) dπ‘₯) = 10, determine the value of ∫_(8)^(βˆ’7) (7𝑔(π‘₯) dπ‘₯).

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Video Transcript

If the integral from negative seven to eight of 𝑔 of π‘₯ with respect to π‘₯ is 10, determine the value of the integral from eight to negative seven of seven 𝑔 of π‘₯ with respect to π‘₯.

Let’s compare these two integrals to see what is different about them. First, we note that the order of the limits has been changed. In the first integral, the lower limit is negative seven. And the upper limit is eight. But in the second integral, the lower limit is eight. And the upper integral is negative seven. We also note that, in the second integral, the function 𝑔 of π‘₯ has been multiplied by seven. So there’re two changes whose effects we’ll need to account for.

Let’s consider this first integral, the integral from negative seven to eight of 𝑔 of π‘₯ with respect to π‘₯. When we integrate, we get a function capital 𝐺 of π‘₯, which we evaluate at eight and negative seven. We subtract the value of this integral at the lower limit of negative seven from the value of this integral at the upper limit of eight. So we have 𝐺 of eight minus 𝐺 of negative seven. Remember, we’re told that the value of this integral is 10. So we have that 10 is equal to 𝐺 of eight minus 𝐺 of negative seven.

Now, let’s consider our second integral. One of the properties of definite integrals is that if we are multiplying by a constant, in this case seven, then that constant can be brought out the front of the integral. So this value of seven here can be brought to the front to multiply the whole integral.

So now, we have seven lots of the integral from eight to negative seven of 𝑔 of π‘₯ with respect to π‘₯. When we integrate, we’ll get that same function capital 𝐺 of π‘₯. But this time, our lower limit is eight and our upper limit is negative seven. So we’ll have seven lots of 𝐺 of negative seven minus 𝐺 of eight.

Now, we can work out what the value inside this bracket is equal to because we know that 𝐺 of eight minus 𝐺 of negative seven is equal to 10. Multiplying this equation through by negative one gives negative 10 is equal to negative 𝐺 of eight plus 𝐺 of negative seven. We could reorder the terms on the right if we wish to give 𝐺 of negative seven minus 𝐺 of eight.

So we now know that the value inside the bracket is negative 10. We therefore have seven multiplied by negative 10, which is equal to negative 70. We’ve used the general property of definite integrals here, which is that if we swap the order of the limits around, so the lower limit becomes the upper limit and vice versa, then the value we get will be the negative of the value we got for the original integral.

Our answer for the value of the integral from eight to negative seven of seven 𝑔 of π‘₯ with respect to π‘₯ is negative 70.

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