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Question Video: Evaluating a Definite Integral Mathematics • Higher Education

Determine ∫_(βˆ’1) ^(1) (βˆ’6π‘₯Β² βˆ’ 3) dπ‘₯.

04:14

Video Transcript

Determine the definite integral from negative one to one of negative six π‘₯ squared minus three with respect to π‘₯.

We’re asked to evaluate the definite integral of a quadratic polynomial. And we know how to evaluate definite integrals by using the fundamental theorem of calculus. So the first thing we’ll do is recall the fundamental theorem of calculus. In fact, we’ll only recall the part which helps us evaluate definite integrals. We know if our integrand lowercase 𝑓 is continuous on the interval of integration the closed interval from π‘Ž to 𝑏 and we have an antiderivative of our integrand, so that’s capital 𝐹 prime of π‘₯ will be equal to lowercase 𝑓 of π‘₯, then the definite integral from π‘Ž to 𝑏 of lowercase 𝑓 of π‘₯ with respect to π‘₯ will be equal to capital 𝐹 evaluated at 𝑏 minus capital 𝐹 evaluated at π‘Ž.

So this gives us a method of evaluating indefinite integrals by using antiderivatives of our integrand. However, the first thing we always need to check is, is our integrand continuous on the interval of integration? In this case, our interval of integration will be the closed interval from negative one to one. So we’ll set our value of π‘Ž equal to negative one and 𝑏 equal to one. Next, our integrand lowercase 𝑓 will be negative six π‘₯ squared minus three. So we need to determine whether negative six π‘₯ squared minus three is continuous on the closed interval from negative one to one. If it is, we can then use the fundamental theorem of calculus to evaluate this integral.

In this case, we see negative six π‘₯ squared minus three is a quadratic polynomial. And we know all polynomials are continuous for all real numbers. So in particular, this means that this polynomial must be continuous on the closed interval from negative one to one. So we can apply the fundamental theorem of calculus to help us evaluate this indefinite integral. This means we need to find an antiderivative of our quadratic.

We actually know several different methods for finding antiderivatives. For example, we could try to reverse the process of differentiation, and this would work. However, we already did this in the general case for polynomials. This gave us the power rule for integration. So we can find an antiderivative of our quadratic by using the power rule for integration. On each term, we want to add one to our exponent and then divide by this new exponent. This gives us negative two π‘₯ cubed minus three π‘₯. And of course we add our constant of integration 𝐢.

And it’s worth reiterating what this means. The derivative of negative two π‘₯ cubed minus three π‘₯ plus 𝐢 with respect to π‘₯ will be equal to our integrand. And this is exactly what it says for capital 𝐹 of π‘₯ to be an antiderivative of lowercase 𝑓 of π‘₯. And this will be an antiderivative for any real value of the constant 𝐢. So we can pick whichever value we want. Normally, we pick zero because it makes the expression the simplest. So we’ll set capital 𝐹 of π‘₯ to be negative two π‘₯ cubed minus three π‘₯.

We’re now ready to apply the fundamental theorem of calculus to evaluate our definite integral. Remember, this tells us that our indefinite integral will be equal to our antiderivative evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration. And we use notation to represent this. We write our antiderivative inside of the square brackets and then our limits of integration outside of this. And all this is is a shorthand notation for writing capital 𝐹 evaluated at one minus capital 𝐹 evaluated at negative one. So all that’s left to do is evaluate this at the limits of integration. We get negative two times one cubed minus three times one minus negative one multiplied by negative one cubed minus three times negative one.

And now we can start simplifying. Inside of our first set of parentheses, we get negative two times one cubed is negative two and negative three times one is negative three. So this simplifies to give us negative two minus three, which is of course equal to negative five. We can do the same to simplify our second set of parentheses. We get negative two minus one cubed is equal to two and negative three multiplied by negative one is equal to three. So our second set of parentheses simplifies to give us two plus three, which is of course equal to five. But remember, we need to subtract this. So we get negative five minus five, which is of course equal to negative 10.

Therefore, by using the fundamental theorem of calculus and the power rule for integration, we were able to show the definite integral from negative one to one of negative six π‘₯ squared minus three with respect to π‘₯ is equal to negative 10.

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