Question

Find the area of the set bounded by the given curves. (a) \(y=x^{2}+9, y=x^{2}-9, x=-1, x=1\) (b) \(y=x+2, y=4-x, x=0\) (c) \(x=y^{2}-4, x=4-y^{2}\) (d) \(y=e^{2 x}, y=-2 x, x=3\)

Short answer

**Short Answer** (a) The area bounded by the given curves is 36 square units. (b) The area bounded by the given curves is 1 square unit. (c) The area bounded by the given curves is 64/3 square units. (d) The area cannot be computed using the definite integral because the given functions do not form a bounded region.

Step-by-step solution

Find the intersections of the functions

First, we need to find the points of intersection between the curves \(y=x^2+9\) and \(y=x^2-9\). To do this, we set these two functions equal to each other and solve for \(x\): \(x^2+9 = x^2-9\) Notice that the \(x^2\) terms cancel out, leaving no solution. This means that these two functions do not intersect. However, the area we are after is actually bounded by the lines \(x=-1\) and \(x=1\). Let's find the limits of integration using these lines.

Set up the integral to find the area

Since the functions do not intersect, we will integrate separately for the functions \(y=x^2+9\) and \(y=x^2-9\) on the interval \([-1, 1]\). The area A will be the difference of the two integrals: \(A = \int_{-1}^1 (x^2+9) dx - \int_{-1}^1 (x^2-9) dx\)

Evaluate the integrals

Evaluate the definite integrals: \(A = \left[\frac{x^3}{3}+9x\right]_{-1}^1 - \left[\frac{x^3}{3}-9x\right]_{-1}^1\) \(A = [ (1/3+9)-(1/3-9) ] - [ (1/3-9)-(1/3+9) ]\) \(A = 18 - (-18) = 36\) The area bounded by the given curves is 36 square units. (b)

Find the intersections of the functions

Find the points of intersection between the lines \(y=x+2\), \(y=4-x\), and \(x=0\) to determine the limits of integration: Setting \(y=x+2\) and \(y=4-x\): \(x+2=4-x\) Solving for x, we get \(x=1\). Hence, the two lines intersect at \((1,3)\).

Set up the integral to find the area

We are looking for the area of the triangle formed by these lines, so we can use the definite integral from \(0\) to \(1\): \(A = \int_{0}^1 (4-x) - (x+2) dx\) Simplify the integrand: \(A = \int_{0}^1 (2-2x) dx\)

Evaluate the integral

Evaluate the definite integral: \(A = [2x-x^2]_{0}^1\) \(A = (2-1)-(0-0) = 1\) The area bounded by the given curves is 1 square unit. (c)

Find the intersections of the functions

Find the points of intersection between the curves \(x=y^2-4\) and \(x=4-y^2\) by setting them equal: \(y^2-4=4-y^2\) Solving for \(y^2\), we get \(y^2=4\). Therefore, \(y=2\) and \(y=-2\).

Set up the integral to find the area

We can set up the definite integral from \(-2\) to \(2\): \(A = \int_{-2}^2 [(4-y^2)-(y^2-4)] dy\) Simplify the integrand: \(A = \int_{-2}^2 (8-2y^2) dy\)

Evaluate the integral

Evaluate the definite integral: \(A = [8y-\frac{2}{3}y^3]_{-2}^2\) \(A = \left[16-\frac{16}{3}\right] - \left[-16+\frac{16}{3}\right]\) \(A = \frac{64}{3}\) The area bounded by the given curves is \(\frac{64}{3}\) square units. (d) In this exercise, one of the curves is given with the exponential function \(y=e^{2x}\), and the other curve is a linear function \(y=-2x\). However, the area we are after is bounded with the additional linear constraint \(x=3\). Unfortunately, these 3 curves do not form a bounded region for a definite integral. Therefore, the area cannot be computed using the definite integral.

Key concepts

Definite Integral

The concept of the definite integral is fundamental in real analysis. It allows us to calculate the accumulation of quantities, such as areas under curves. When dealing with functions and curves, the definite integral can be thought of as the signed area between the curve and the x-axis over an interval.

For a function \( f(x) \), the definite integral from \( a \) to \( b \), represented as \( \int_{a}^{b} f(x) \, dx \), gives us the net area between the curve \( f(x) \) and the x-axis from \( x = a \) to \( x = b \). This includes areas above the x-axis as positive and areas below as negative.

• The limits of integration, \( a \) and \( b \), define the interval over which we are calculating the area.
• If \( f(x) \) is positive in the interval, the integral gives the area above the x-axis.
• If \( f(x) \) dips below the x-axis, those areas are subtracted from the total.

Understanding this helps in solving problems related to the "area under curves," where integration techniques are used to find the precise area within bounds established by intersections or specific limits along the x-axis.

Area under a Curve

To find the area under a curve is to compute the definite integral of a function over a specific interval. This can represent physical quantities, like distance traveled or the accumulated quantity over time, depending on the context of the function.

In mathematical terms, the area under the curve of \( y = f(x) \) between two points, \( x = a \) and \( x = b \), is calculated using the integral \( \int_{a}^{b} f(x) \, dx \). This process essentially sums an infinite number of infinitesimally small rectangles under the curve from \( a \) to \( b \).

• To ensure all areas are positive, it's often necessary to take separate integrals for portions of the function that are above and below the x-axis, especially when they involve more complex shapes or multiple bounds.
• Using symmetry can simplify calculations; if a curve is symmetric about an axis, integrals may be computed over half the interval and then doubled.

Utilizing the area under a curve is instrumental in solving problems about displacement and area, providing a precise measure that can be applied to real-world scenarios in physics and engineering.

Intersection of Curves

Intersection of curves is a critical concept when finding the area between curves. It defines the points at which two or more curves meet, and these points often serve as limits of integration for definite integrals.

To find intersections, you set equations of the curves equal to each other and solve for the relevant variable (often \( x \) or \( y \)), which delivers the coordinates at which curves cross. These intersection points are essential because they mark the boundaries of the region whose area you wish to calculate.

• For example, to find where \( y = x^2 + 9 \) intersects \( y = x^2 - 9 \), you would set \( x^2 + 9 = x^2 - 9 \), though in this case, note they do not intersect, and boundaries must be otherwise determined.
• Successful intersection analysis might reveal multiple crossover points requiring separate integrals over specified intervals when calculating areas between the curves.
• For calculus applications, if solving for \( y \) yields complex values or no real solutions, it may indicate the curves only meet graphically but not in the real plane.

Understanding intersection points is crucial in applying real analysis for measuring enclosed regions accurately.