Question

You should solve the following problems without using a graphing calculator. True or False The area of the region enclosed by the graph of $$y=x^{2}+1$$ and the line $$y=10$$ is $$36 .$$ Justify your answer.

Short answer

Yes, the area of the region enclosed is indeed \(36\).

Step-by-step solution

Expression for Area

Area is the integral of the function from the lower boundary to the upper boundary. In this case, we have to find the bounds where the given graph \(y=x^{2}+1\) and line \(y=10\) intersect. Solving the equations \(x^{2}+1 = 10\) gives the intersections at \(x=-3\) and \(x=3\). Therefore, Area, A = \(\int_{-3}^{3} (10 - (x^{2}+1)) dx\).

Evaluating the Definite Integral

Evaluate the integral, using the fundamental theorem of calculus. So, \(A=\int_{-3}^{3} (9 - x^{2}) dx\) = \([9x - x^{3}/3]_{-3}^{3}\).

Substitute the bounds and simplify

Substitute the bounds into the antiderivative to obtain the area. \(A=[27 - 9] - [-27 - 9] = 36\).

Key concepts

Definite Integral

When we talk about finding the area between two curves in calculus, the definitive tool we use is the definite integral. A definite integral calculates the accumulation of quantities, which can be areas under a curve, between a range of values, typically along the x-axis.

To understand this concept better, let's consider an example where we want to find the area between the graph of a quadratic function, say \( y = x^2 \), and a straight line, let's choose \( y = 4 \), between the points where they intersect. To find this area, we would integrate the difference of the functions (top function minus bottom function) from the left intersection point to the right intersection point.

Thus, the integral would be expressed as \( \text{Area} = \text{integrate from left intersection point to right intersection point} (y_{\text{top curve}} - y_{\text{bottom curve}}) \text{dx} \). The result of this calculation gives us the area between the curves for that specific range.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a principal theorem that creates a connection between differentiation and integration, two main concepts in calculus. This theorem states that if a function is continuous over an interval and has an antiderivative within that interval, then the integral of the function over that interval can be calculated by using its antiderivatives.

To apply this theorem, we first find the antiderivative of the function being integrated. Remember, an antiderivative is a function whose derivative is the original function. Once we have the antiderivative, we evaluate it at the upper and lower bounds of the interval and subtract the two results. This gives us the exact area under the curve between the two points. It can be symbolically represented as: \( \text{A}=\text{F(b)} - \text{F(a)} \) where F is the antiderivative, and a and b are the bounds of the interval.

Intersection of Graphs

When solving area problems between two curves, identifying the points of intersection is crucial. These are the points where the graphs of the functions cross each other. The x-values of these intersection points become the limits of integration for the definite integral.

To find the intersection points, we set the equations of the two curves equal to each other and solve for x. For instance, to find where the graph of \( y = x^2 + 1 \) intersects with the line \( y = 10 \), we would solve the equation \( x^2 + 1 = 10 \). The solutions to this equation give us the x-values where the curves intersect, and these are the values we use as the limits when setting up the definite integral.

Antiderivatives

An antiderivative, also known as an indefinite integral, is essentially the reverse operation of differentiation. If we know the derivative of a function, we can work backwards to find the original function, which is the antiderivative. In the context of finding areas, the antiderivative gives us a function that we can evaluate at specific points to calculate the accumulated area under a curve.

To determine the area between curves using antiderivatives, we find the antiderivative of the function that defines the area. In the earlier mentioned exercise, the function is \( 9 - x^2 \), and its antiderivative is \( 9x - \frac{x^3}{3} \). By evaluating this antiderivative at the intersection points, which serve as the bounds of integration, and taking the difference, we obtain the total area enclosed between the curves.