Area Between Two Curves

1) If f(x) ≥ g(x) on [a, b], which formula correctly represents the area between y = f(x) and y = g(x)?

A = ∫[a to b] [f(x) + g(x)] dx

A = ∫[a to b] [f(x) − g(x)] dx

A = ∫[a to b] [g(x) − f(x)] dx

A = ∫[a to b] [f(x) · g(x)] dx

To find the area between the curves y = f(x) and y = g(x) where f(x) is above g(x) on the interval [a, b], we need to compute the vertical distance between the two functions. This distance is given by f(x) - g(x). Integrating this difference from a to b captures the total area between the curves, as it sums the heights of the vertical slices across the interval. Thus, the formula A = ∫[a to b] [f(x) − g(x)] dx accurately represents the area between the two functions.

Explanation

To find the area between the curves y = f(x) and y = g(x) where f(x) is above g(x) on the interval [a, b], we need to compute the vertical distance between the two functions. This distance is given by f(x) - g(x). Integrating this difference from a to b captures the total area between the curves, as it sums the heights of the vertical slices across the interval. Thus, the formula A = ∫[a to b] [f(x) − g(x)] dx accurately represents the area between the two functions.

2) When computing the area between two curves expressed as x = f(y) and x = g(y) where f(y) ≥ g(y) on [c, d], the correct integral setup is:

A = ∫[c to d] [f(y) − g(y)] dy, integrating with respect to y

A = ∫[c to d] [f(x) − g(x)] dx, integrating with respect to x

A = ∫[c to d] [g(y) − f(y)] dy, integrating with respect to y

A = ∫[c to d] [f(y) + g(y)] dy, integrating with respect to y

To find the area between two curves defined as x = f(y) and x = g(y), where f(y) is greater than or equal to g(y) over the interval [c, d], we need to calculate the vertical distance between the curves. This distance is represented by the difference f(y) - g(y). By integrating this difference with respect to y from c to d, we accurately capture the area between the two curves across the specified interval. Thus, the integral setup reflects the correct approach for determining the area.

Explanation

To find the area between two curves defined as x = f(y) and x = g(y), where f(y) is greater than or equal to g(y) over the interval [c, d], we need to calculate the vertical distance between the curves. This distance is represented by the difference f(y) - g(y). By integrating this difference with respect to y from c to d, we accurately capture the area between the two curves across the specified interval. Thus, the integral setup reflects the correct approach for determining the area.

3) In Example 1, the area between f(x) = √x and g(x) = x² is found by first solving √x = x². What are the intersection points?

X = 0 and x = 2

X = −1 and x = 1

X = 0 and x = 1

X = 1 and x = 4

To find the intersection points of the functions f(x) = √x and g(x) = x², we set them equal: √x = x². Squaring both sides gives x = x^4, which simplifies to x^4 - x = 0. Factoring this equation yields x(x^3 - 1) = 0, leading to solutions x = 0 and x = 1. These points represent the x-values where the two functions intersect, confirming that the area between them can be calculated between x = 0 and x = 1.

Explanation

To find the intersection points of the functions f(x) = √x and g(x) = x², we set them equal: √x = x². Squaring both sides gives x = x^4, which simplifies to x^4 - x = 0. Factoring this equation yields x(x^3 - 1) = 0, leading to solutions x = 0 and x = 1. These points represent the x-values where the two functions intersect, confirming that the area between them can be calculated between x = 0 and x = 1.

4) Using Example 1, evaluate A = ∫[0 to 1] [√x − x²] dx. What is the exact area?

1/2

1/6

2/3

1/3

To evaluate the integral A = ∫[0 to 1] (√x - x²) dx, we first find the antiderivative of the integrand. The antiderivative of √x is (2/3)x^(3/2), and the antiderivative of x² is (1/3)x³. We then compute the definite integral by evaluating the antiderivative from 0 to 1:

A = [(2/3)(1) - (1/3)(1)] - [(2/3)(0) - (1/3)(0)] = (2/3 - 1/3) = 1/3.

Thus, the exact area under the curve from 0 to 1 is 1/3.

Explanation

To evaluate the integral A = ∫[0 to 1] (√x - x²) dx, we first find the antiderivative of the integrand. The antiderivative of √x is (2/3)x^(3/2), and the antiderivative of x² is (1/3)x³. We then compute the definite integral by evaluating the antiderivative from 0 to 1:

A = [(2/3)(1) - (1/3)(1)] - [(2/3)(0) - (1/3)(0)] = (2/3 - 1/3) = 1/3.

Thus, the exact area under the curve from 0 to 1 is 1/3.

5) In Example 3, the area between x = −y² + 10 and x = (y − 2)² is computed. What are the y-values of the intersection points?

Y = 0 and y = 4

Y = −1 and y = 3

Y = 1 and y = 5

Y = −2 and y = 2

To find the intersection points of the curves \( x = -y^2 + 10 \) and \( x = (y - 2)^2 \), we set them equal to each other:

\[
-y^2 + 10 = (y - 2)^2
\]

Expanding and rearranging gives a quadratic equation. Solving this equation yields the y-values where the two curves intersect. The solutions are \( y = -1 \) and \( y = 3 \), indicating these are the points where the area between the curves is defined.

Explanation

To find the intersection points of the curves \( x = -y^2 + 10 \) and \( x = (y - 2)^2 \), we set them equal to each other:

\[
-y^2 + 10 = (y - 2)^2
\]

Expanding and rearranging gives a quadratic equation. Solving this equation yields the y-values where the two curves intersect. The solutions are \( y = -1 \) and \( y = 3 \), indicating these are the points where the area between the curves is defined.

6) In Example 3, after simplifying the integrand [−y² + 10 − (y − 2)²], the integral becomes A = ∫[−1 to 3] [−2y² + 4y + 6] dy. What is the final area?

32/3

64/3

128/3

16

48

To find the final area, we first simplify the integrand, resulting in the expression \(-2y² + 4y + 6\). We then compute the definite integral of this polynomial from \(-1\) to \(3\). Evaluating the integral involves finding the antiderivative, substituting the limits, and calculating the difference. After performing the integration and simplification, we find that the area under the curve over the specified interval equals \(64/3\). This represents the total area bounded by the curve and the x-axis within the limits of integration.

Explanation

To find the final area, we first simplify the integrand, resulting in the expression \(-2y² + 4y + 6\). We then compute the definite integral of this polynomial from \(-1\) to \(3\). Evaluating the integral involves finding the antiderivative, substituting the limits, and calculating the difference. After performing the integration and simplification, we find that the area under the curve over the specified interval equals \(64/3\). This represents the total area bounded by the curve and the x-axis within the limits of integration.

7) In Example 2, the area between y = 2x² + 10 and y = 4x + 16 over [−2, 5] requires splitting into subregions. What are the x-coordinates where the two curves intersect?

X = 0 and x = 2

X = −2 and x = 5

X = −1 and x = 3

X = 1 and x = 4

To find the x-coordinates where the curves \( y = 2x^2 + 10 \) and \( y = 4x + 16 \) intersect, we need to set the equations equal to each other and solve for \( x \). This involves rearranging the equation to form a quadratic equation and applying the quadratic formula or factoring. After solving, the points of intersection are found to be at \( x = -1 \) and \( x = 3 \). These points indicate where the area between the curves needs to be calculated, hence the need for splitting the region.

Explanation

To find the x-coordinates where the curves \( y = 2x^2 + 10 \) and \( y = 4x + 16 \) intersect, we need to set the equations equal to each other and solve for \( x \). This involves rearranging the equation to form a quadratic equation and applying the quadratic formula or factoring. After solving, the points of intersection are found to be at \( x = -1 \) and \( x = 3 \). These points indicate where the area between the curves needs to be calculated, hence the need for splitting the region.

8) In Example 2, the total area A = A₁ + A₂ + A₃ is computed over three subintervals. What is the total area?

128/3

64/3

142/3

150/3

To find the total area A over three subintervals, we sum the areas calculated for each subinterval (A₁, A₂, and A₃). In this case, the individual areas, when added together, yield a total area of 142/3. This result indicates that the calculations for each subinterval were performed correctly, and their combination reflects the overall area under the curve or the total area being analyzed.

Explanation

To find the total area A over three subintervals, we sum the areas calculated for each subinterval (A₁, A₂, and A₃). In this case, the individual areas, when added together, yield a total area of 142/3. This result indicates that the calculations for each subinterval were performed correctly, and their combination reflects the overall area under the curve or the total area being analyzed.

Area Between Two Curves

1) If f(x) ≥ g(x) on [a, b], which formula correctly represents the area between y = f(x) and y = g(x)?

2)

What first name or nickname would you like us to use?

2) When computing the area between two curves expressed as x = f(y) and x = g(y) where f(y) ≥ g(y) on [c, d], the correct integral setup is:

3) In Example 1, the area between f(x) = √x and g(x) = x² is found by first solving √x = x². What are the intersection points?

4) Using Example 1, evaluate A = ∫[0 to 1] [√x − x²] dx. What is the exact area?

5) In Example 3, the area between x = −y² + 10 and x = (y − 2)² is computed. What are the y-values of the intersection points?

6) In Example 3, after simplifying the integrand [−y² + 10 − (y − 2)²], the integral becomes A = ∫[−1 to 3] [−2y² + 4y + 6] dy. What is the final area?

7) In Example 2, the area between y = 2x² + 10 and y = 4x + 16 over [−2, 5] requires splitting into subregions. What are the x-coordinates where the two curves intersect?

8) In Example 2, the total area A = A₁ + A₂ + A₃ is computed over three subintervals. What is the total area?