### Video Transcript

Evaluate the definite integral from zero to one of four plus five ๐ก multiplied by the square root of ๐ก with respect to ๐ก.

In this question, weโre asked to evaluate a definite integral. And we know we can try evaluating definite integrals by using the fundamental theorem of calculus. So weโll start by recalling part of the fundamental theorem of calculus. we know if lowercase ๐ is continuous on the closed interval from ๐ to ๐ and capital ๐น prime of ๐ฅ is equal to lowercase ๐ of ๐ฅ, then the definite integral from ๐ to ๐ of lowercase ๐ of ๐ฅ with respect to ๐ฅ is equal to capital ๐น evaluated at ๐ minus capital ๐น evaluated at ๐.

In other words, if our integrand is continuous on the interval of integration and we can find an antiderivative of our integrand, then we can evaluate our definite integral by evaluating our antiderivative. So we should start by looking at the definite integral given to us in the question. First, the lower limit of integration is zero and the upper limit of integration is one. So weโll set ๐ equal to zero and ๐ equal to one. Next, we can see weโre integrating a function in ๐ก with respect to ๐ก. So we should rewrite our fundamental theorem of calculus to be about the variable ๐ก.

Then once weโve done this, we can set lowercase ๐ of ๐ก to be our integrand. Now to use the fundamental theorem of calculus, thereโs two things weโre going to need. First, weโre going to need to show that our integrand lowercase ๐ is continuous on the interval of integration, the closed interval from zero to one. Then once weโve done that, weโre also going to need to find an antiderivative of this integrand. Letโs start with checking the continuity of our integrand. We can see that our integrand is the product of two functions. Itโs a linear function, four plus five ๐ก multiplied by the square root of ๐ก.

And in fact, we know both of these are continuous across their entire domain. We know four plus five ๐ก is a polynomial, so itโs continuous for all real values of ๐ก. And the square root of ๐ก is continuous. However, its domain is only for values of ๐ก greater than or equal to zero. So our integrand is the product of two continuous functions. This means it will be continuous across its entire domain. In this case, thatโs all values of ๐ก greater than or equal to zero. So in particular, itโs continuous on the closed interval from zero to one.

So now that weโve shown our integrand is continuous on the interval of integration, all we need to do now is find our antiderivative. And the easiest way to find an antiderivative of a function is to use what we know about indefinite integrals. Letโs try and find the indefinite integral of four plus five ๐ก multiplied by the square root of ๐ก with respect to ๐ก. And we can do this by using our laws of exponents and the power rule for integration.

First, weโll rewrite the square root of ๐ก as ๐ก to the power of one-half. Now, we want to distribute ๐ก to the power of one-half over our parentheses by using our laws of exponents. Doing this gives us the indefinite integral of four ๐ก to the power of one-half plus five ๐ก to the power of three over two with respect to ๐ก. And we know how to integrate this by using the power rule for integration. We want to do this term by term, adding one to our exponents of ๐ก and then dividing by the new exponents of ๐ก.

Applying this to our first term, we get four ๐ก to the power of three over two all divided by three over two. And applying this to our second term, we get five ๐ก to the power of five over two divided by five over two. And remember, we need to add a constant of integration ๐ถ. So this gives us the general antiderivative of our integrand. And in fact, we can simplify slightly. Instead of dividing by our fractions, weโll multiply by the reciprocals. So by multiplying by the reciprocals of our denominator and then simplifying, we get eight ๐ก to the power of three over two divided by three plus two ๐ก to the power of five over two plus ๐ถ.

And remember, this is the general antiderivative of our integrand. It will be an antiderivative for any value of ๐ถ. But we donโt need the general antiderivative. We just need any specific antiderivative. So we can choose our value of ๐ถ. Weโll pick ๐ถ is equal to zero because this makes it easier. Now that weโve shown the top integrand is continuous in the interval of integration and we found our antiderivative, weโre ready to evaluate our definite integral. By using the fundamental theorem of calculus, we can evaluate our indefinite integral by just evaluating our antiderivative at the upper and lower limits of integration.

We represent this in the following notation. We write our antiderivative inside of our square brackets and then write our limits of integration outside. So all thatโs left to do now is evaluate this at the limits of integration. Substituting ๐ก is equal to one into our antiderivative, we get eight times one to the power of three over two divided by three plus two times one to the power of five over two. Then we want to evaluate our antiderivative at zero. Remember, we need to subtract this. This means weโre subtracting eight times zero to the power of three over two divided by three plus two times zero to the power of five over two.

And now, we can start simplifying. First, one to the power of three over two and one to the power of five over two are just equal to one. Next, zero to the power of three over two and zero to the power of five over two are equal to zero. So everything inside of our parentheses has a factor of zero, so this simplifies to give us zero. So this simplifies to give us eight over three plus two, which we can calculate is our final answer of 14 over three.

Therefore, by using the fundamental theorem of calculus and the power rule for integration, we were able to show the definite integral from zero to one of four plus five ๐ก all multiplied by the square root of ๐ก with respect to ๐ก is equal to 14 over three.