Question Video: Evaluating the Definite Integral of a Root Function | Nagwa Question Video: Evaluating the Definite Integral of a Root Function | Nagwa

# Question Video: Evaluating the Definite Integral of a Root Function Mathematics • Third Year of Secondary School

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Evaluate โซ_(16)^(18) โ(3/๐ง) d๐ง.

05:02

### Video Transcript

Evaluate the definite integral from 16 to 18 of the square root of three over ๐ง with respect to ๐ง.

In this question, weโre asked to evaluate a definite integral, and we know two different ways of doing this. We could try and find the area under the curve defined by this equation between ๐ง is equal to 16 and ๐ง is equal to 18. However, we wouldnโt know how to do this directly. So instead, weโre going to use the fundamental theorem of calculus. We recall this tells us if lowercase ๐ is continuous on the closed interval from ๐ to ๐ and capital ๐น prime of ๐ง is equal to lowercase ๐ of ๐ง, then the definite integral from ๐ to ๐ of lowercase ๐ of ๐ง with respect to ๐ง is equal to capital ๐น of ๐ minus capital ๐น of ๐.

In other words, if our integrand is continuous on the domain of integration and we can find the antiderivative of our integrand, we can just use this to evaluate our definite integral. So, letโs start doing this for the integral given to us in the question. First, the lower limit of integration is 16 and the upper limit is 18, so weโll set ๐ equal to 16 and ๐ equal to 18. Next, weโll set our integrand, which is the square root of three over ๐ง, as our function lowercase ๐ of ๐ง.

To use the fundamental theorem of calculus, we need to check the top integrand is continuous on the interval of integration. In this case, thatโs the closed interval from 16 to 18. To do this, weโre going to need to take a closer look at our integrand. We can see itโs the composition of two functions. Weโre taking the square root of three divided by ๐ง. And we know a lot about these two functions. For example, three over ๐ง is a rational function. So, it will be continuous across its entire domain, which is everywhere except where ๐ง is equal to zero. And we also know the square root function is continuous across its entire domain. So, our integrand is the composition of two continuous functions, which means it will be continuous on its entire domain.

So to check that our integrand is continuous on the closed interval from 16 to 18, we just need to check itโs defined for all values of ๐ง between 16 and 18 inclusive. And in fact, we could just see this by substituting these values into our integrand. If we substitute any value of ๐ง between 16 and 18, weโre just taking the square root of a positive number, and we know this is well defined. So, our integrand is defined for all values of ๐ง on this interval. And we know because itโs the composition of two continuous functions. This means itโs also continuous on this interval.

So, we can use the fundamental theorem of calculus to evaluate this integral. We just need to find our antiderivative. And the easiest way to find the antiderivative of our integrand will be to use what we know about indefinite integrals. We know the indefinite integral of the square root of three over ๐ง with respect to ๐ง will give us the general antiderivative of our integrand. And to evaluate this indefinite integral, weโre going to use our rules for exponents.

First, we know the square root of three over ๐ง is equal to the square root of three divided by the square root of ๐ง. But then, we can use another one of our laws of exponents. We can rewrite this as the square root of three multiplied by ๐ง to the power of negative one-half. And now that weโve written our integrand in this form, we can see we can evaluate this by using the power rule for integration. We want to add one to our exponent of ๐ง and then divide by this new exponent. This gives us root three times ๐ง to the power of one-half all divided by one-half. And before we add our constant of integration, we can simplify this. Weโll multiply both our numerator and our denominator through by two. This gives us two root three times ๐ง to the power of one-half. And weโll add our constant of integration ๐ถ.

And finally, by using our laws of exponents, weโll rewrite ๐ง to the power of one-half as the square root of ๐ง. So, this gives us two root three times root ๐ง plus ๐ which is our general antiderivative of our integrand. This will be an antiderivative for any value of ๐, but we only need one antiderivative. So, weโll choose ๐ is equal to zero. This gives us capital ๐น of ๐ง is equal to two times root three times root ๐ง. And now that we found our antiderivative, weโre ready to use the fundamental theorem of calculus to evaluate our definite integral.

We know itโs equal to capital ๐น evaluated at 18 minus capital ๐น evaluated at 16. We represent this in the following notation. So all thatโs left to do now is evaluate this at the limits of integration. We have that capital ๐น evaluated at 18 minus capital ๐น evaluated at 16 is equal to two root three root 18 minus two root three root 16. And we can simplify both of these expressions. First, root 16 is equal to four. Next, we know that 18 is equal to three squared times two. So, the square root of 18 can be simplified to give us three root two.

And now, weโre on the last bit of simplification. First, two times three is equal to six. Next, root three times root two is equal to root six. Finally, in our second term, two times four is equal to eight. The last thing weโll do is reorder these two terms to give us our final answer of negative eight root three plus six root six.

Therefore, by using the fundamental theorem of calculus and the power rule for integration, we were able to show the definite integral from 16 to 18 of the square root of three over ๐ง with respect to ๐ง is equal to negative eight root three plus six root six.

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