Question

(a) Compute the area under the parabola \(\mathrm{y}=2 \mathrm{x}^{2}\) on the interval \([1,2]\) as the limit of a sum. (b) Let \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{2}\) and note that \(\mathrm{g}(\mathrm{x})=\frac{2}{3} \mathrm{x}^{3}\) defines a function that satisfies \(g^{\prime}(\mathrm{x})=\mathrm{f}(\mathrm{x})\) on the interval [1,2] Verify that the area computed in part (a) satisfies \(\mathrm{A}=\mathrm{g}(2)-\mathrm{g}(\mathrm{l})\) (c) The function defined by \(h(x)=\frac{2}{3} x^{3}+C\) for any constant \(\mathrm{C}\) also satisfies \(\mathrm{h}^{\prime}(\mathrm{x})=\mathrm{f}(\mathrm{x})\). Is it true that the area in part (a) satisfies \(\mathrm{A}=\mathrm{h}(2)-\) \(\mathrm{h}(1) ?\)

Short answer

To summarize the solution, we first computed the area under the parabola \(y=2x^2\) on the interval \([1,2]\) using the limit of a sum and found the area to be \(\frac{14}{3}\). We then verified that the area satisfies the equation \(A = g(2) - g(1)\) for the given function \(g(x) = \frac{2}{3}x^3\). Finally, we determined that the area also satisfies the equation \(A = h(2) - h(1)\) for the function \(h(x) = \frac{2}{3}x^3 + C\), regardless of the value of the constant \(C\).

Step-by-step solution

(a) Compute the area using the limit of a sum

We will divide the interval \([1,2]\) into \(n\) equal rectangles of width \(\Delta x = \frac{2-1}{n}\). Let's label the right-hand endpoints of these rectangles as \(x_i = 1 + i\Delta x = 1 + \frac{i}{n}\). Now, the height of the \(i\)-th rectangle will be given by the function \(f(x)=2x^2\). Its area will be: \[f(x_i)\Delta x = 2(1 + \frac{i}{n})^2 \cdot \frac{1}{n}\] Now, summing the areas of all rectangles, we get: \[\sum_{i=1}^n 2(1 + \frac{i}{n})^2 \cdot \frac{1}{n}\] Finally, taking the limit as the number of rectangles \(n\) approaches infinity gives us the exact area under the curve: \[A = \lim_{n \to \infty} \sum_{i=1}^n 2(1 + \frac{i}{n})^2 \cdot \frac{1}{n} = \int_1^2 2x^2 dx\]

(a) Evaluate the integral to find the area

By evaluating the integral, we can find the value of the area under the curve: \[A = \int_1^2 2x^2 dx = \Big[\frac{2}{3}x^3\Big]_1^2 = \frac{2}{3}(2^3 - 1^3) = \frac{2}{3}(8 - 1) = \frac{14}{3}\]

(b) Verify that the area satisfies A=g(2)-g(1)

We are given the function \(g(x) = \frac{2}{3}x^3\). We will check if the area we found in part (a) satisfies the equation \(A = g(2) - g(1)\): \[A = g(2) - g(1) = \frac{2}{3}(2^3) - \frac{2}{3}(1^3) = \frac{14}{3}\] Thus, the area computed in part (a) satisfies the given equation.

(c) Determine if A=h(2)-h(1) holds for h(x)=\frac{2}{3}x^3+C

We are asked to determine if the area in part (a) satisfies \(A = h(2) - h(1)\) for the function \(h(x) = \frac{2}{3}x^3 + C\): \[A = h(2) - h(1) = \big(\frac{2}{3}(2^3) + C\big) - \big(\frac{2}{3}(1^3) + C\big) = \frac{14}{3}\] Thus, the area in part (a) satisfies the equation \(A = h(2) - h(1)\) for any constant \(C\).

Key concepts

Area Under the Curve

The term area under the curve in calculus refers to the space between a curve on a graph and the x-axis. Calculating this area is fundamental in mathematics as it represents quantities such as distance traveled over time, the amount of substance used, and other instances where accumulation is studied. One method to estimate this area is by dividing the region into simple shapes, for which the area can be easily calculated, and then summing these areas. This approach can lead to exact calculations when using integrals.

For example, the exercise involves finding the area under the curve of the function f(x) = 2x². By integrating this function from x = 1 to x = 2, we can find the exact value of the area, which is a primary concept in definite integral calculus.

Limit of a Sum

The concept of the limit of a sum lays the foundation for understanding integrals as the area under a curve. This idea starts by approximating the area using a sum of the areas of rectangles under the curve. Each rectangle has a height determined by the value of the function at a chosen point and a width which is a fragment of the interval. As the number of rectangles increases (thus their width decreases), the approximation becomes more accurate. By taking the limit as the number of rectangles approaches infinity, we achieve the exact area, transitioning from a discrete sum to a definite integral. This process, demonstrated in the exercise, involves using Riemann sums to calculate the area under f(x) = 2x² over the interval [1,2].

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation and integration, showing that they are, in a sense, inverse processes. This theorem has two parts: the first part guarantees that if a function is continuous over an interval, then its integral is differentiable and the derivative is the original function. The second part states that if you have an antiderivative of a function, you can calculate definite integrals (or the exact area under the curve) over an interval by evaluating the difference of the antiderivative at the endpoints of the interval. In the given exercise, this theorem is exemplified by the fact that the area under f(x) from x = 1 to x = 2 is represented by A = g(2) - g(1), with g(x) being an antiderivative of f(x).

Antiderivatives

An antiderivative of a function f(x), also known as an indefinite integral, is a function whose derivative is f(x). One key feature of antiderivatives is that they are not unique, since adding any constant to an antiderivative results in another valid antiderivative. This is why a family of antiderivatives is often written as G(x) = F(x) + C, where C is any constant. In the exercise, we confirm that even with any constant added to the antiderivative h(x) = (2/3)x³ + C, the area under the curve between x = 1 and x = 2 remains constant, as the constant C cancels out, confirming the invariance of the definite integral to the addition of a constant to the antiderivative.