Video: Finding the Area under the Curve of a Quadratic Function

Let 𝑓(π‘₯) = 2π‘₯Β² + 3. Determine the area bounded by the curve 𝑦 = 𝑓(π‘₯), the π‘₯-axis, and the two lines π‘₯ = βˆ’1 and π‘₯ = 5.

02:21

Video Transcript

Let 𝑓 of π‘₯ equal two π‘₯ squared plus three. Determine the area bounded by the curve 𝑦 equals 𝑓 of π‘₯, the π‘₯-axis, and the two lines π‘₯ equals negative one and π‘₯ equals five.

Let’s begin with a sketch of the region whose area we’re looking to calculate. It’s bounded by a quadratic curve with a positive leading coefficient and a 𝑦-intercept of three. It’s also bounded by the two vertical lines π‘₯ equals negative one and π‘₯ equals five and the π‘₯-axis. So it’s this area here that we’re looking to find.

We recall that the area bounded by the curve 𝑦 equals 𝑓 of π‘₯, the π‘₯-axis, and the two vertical lines π‘₯ equals π‘Ž and π‘₯ equals 𝑏 can be found by evaluating the definite integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ with respect to π‘₯. Our function 𝑓 of π‘₯ is two π‘₯ squared plus three. The lower limit for our integral, the value of π‘Ž, is the lower value of π‘₯. That’s negative one. And the upper limit, the value of 𝑏, is the upper limit of π‘₯. That’s five. So the area we’re looking for is equal to the integral from negative one to five of two π‘₯ squared plus three with respect to π‘₯.

We recall that, in order to integrate powers of π‘₯ not equal to negative one, we increase the power by one and then divide by the new power. So the integral of two π‘₯ squared is two π‘₯ cubed over three, and the integral of three is three π‘₯. We have then that the area is equal to two π‘₯ cubed over three plus three π‘₯ evaluated between negative one and five.

Remember, there’s no need for a constant of integration here, as this is a definite integral. We then substitute the limits, giving two multiplied by five cubed over three plus three multiplied by five minus two multiplied by negative one cubed over three plus three multiplied by negative one. That’s 250 over three plus 15 minus negative two-thirds minus three. That simplifies to 102. And so we can say that the area of the region bounded by the curve 𝑦 equals 𝑓 of π‘₯, the π‘₯-axis, and the two lines π‘₯ equals negative one and π‘₯ equals five found by evaluating the definite integral of our function 𝑓 of π‘₯ between the limits of negative one and five is 102 square units.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.