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

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

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Evaluate โซ_(0)^(1) (4 + 5๐ก) โ(๐ก) d๐ก.

05:14

### Video Transcript

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

In this question, weโre asked to evaluate a definite integral. And we know we can try evaluating definite integrals by using the fundamental theorem of calculus. So weโll 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 ๐น prime of ๐ฅ is equal to lowercase ๐ of ๐ฅ, then the definite integral from ๐ to ๐ of lowercase ๐ of ๐ฅ with respect to ๐ฅ is equal to capital ๐น evaluated at ๐ minus capital ๐น 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 weโll set ๐ equal to zero and ๐ equal to one. Next, we can see weโre integrating a function in ๐ก with respect to ๐ก. So we should rewrite our fundamental theorem of calculus to be about the variable ๐ก.

Then once weโve done this, we can set lowercase ๐ of ๐ก to be our integrand. Now to use the fundamental theorem of calculus, thereโs two things weโre going to need. First, weโre 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 weโve done that, weโre also going to need to find an antiderivative of this integrand. Letโs start with checking the continuity of our integrand. We can see that our integrand is the product of two functions. Itโs a linear function, four plus five ๐ก multiplied by the square root of ๐ก.

And in fact, we know both of these are continuous across their entire domain. We know four plus five ๐ก is a polynomial, so itโs continuous for all real values of ๐ก. And the square root of ๐ก is continuous. However, its domain is only for values of ๐ก 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, thatโs all values of ๐ก greater than or equal to zero. So in particular, itโs continuous on the closed interval from zero to one.

So now that weโve 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. Letโs try and find the indefinite integral of four plus five ๐ก multiplied by the square root of ๐ก with respect to ๐ก. And we can do this by using our laws of exponents and the power rule for integration.

First, weโll rewrite the square root of ๐ก as ๐ก to the power of one-half. Now, we want to distribute ๐ก to the power of one-half over our parentheses by using our laws of exponents. Doing this gives us the indefinite integral of four ๐ก to the power of one-half plus five ๐ก to the power of three over two with respect to ๐ก. 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 ๐ก and then dividing by the new exponents of ๐ก.

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

And remember, this is the general antiderivative of our integrand. It will be an antiderivative for any value of ๐ถ. But we donโt need the general antiderivative. We just need any specific antiderivative. So we can choose our value of ๐ถ. Weโll pick ๐ถ is equal to zero because this makes it easier. Now that weโve shown the top integrand is continuous in the interval of integration and we found our antiderivative, weโre 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 thatโs left to do now is evaluate this at the limits of integration. Substituting ๐ก 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 weโre 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 ๐ก all multiplied by the square root of ๐ก with respect to ๐ก is equal to 14 over three.

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