Problem 23 - Thomas Calculus

Question

Evaluate \(\int_{C} x y d x+(x+y) d y\) along the curve \(y=x^{2}\) from \((-1,1)\) to \((2,4) .\)

Step-by-Step Solution

Answer

The value of the line integral is \( \frac{207}{12} \).

Step 1: Parameterize the Curve

The curve is given by the equation \( y = x^2 \), so we can parameterize it as \( \mathbf{r}(t) = (t, t^2) \) with \( t \) ranging from -1 to 2. This means that \( x = t \) and \( y = t^2 \).

Step 2: Express the Line Integral

Rewrite the integral \( \int_{C} x y \, dx + (x+y) \, dy \) using the parameterization: substitute \( x = t \) and \( y = t^2 \). Thus, \( dx = dt \) and \( dy = 2t \, dt \). The integral becomes \( \int_{-1}^{2} (t)(t^2) \, dt + (t + t^2)(2t \, dt) \).

Step 3: Simplify the Integrand

Simplify the expression inside the integral: \( t(t^2) = t^3 \) and \( (t + t^2)(2t) = 2t^2 + 2t^3 \). The integrand simplifies to \( t^3 + 2t^2 + 2t^3 = 3t^3 + 2t^2 \).

Step 4: Evaluate the Definite Integral

Now, calculate the integral: \[ \int_{-1}^{2} (3t^3 + 2t^2) \, dt. \] First, integrate each term: \( \int 3t^3 \, dt = \frac{3}{4}t^4 \) and \( \int 2t^2 \, dt = \frac{2}{3}t^3 \). Combining, the antiderivative is \( \frac{3}{4}t^4 + \frac{2}{3}t^3 \).

Step 5: Apply the Limits

Evaluate the antiderivative from -1 to 2: \[ \left( \frac{3}{4}(2)^4 + \frac{2}{3}(2)^3 \right) - \left( \frac{3}{4}(-1)^4 + \frac{2}{3}(-1)^3 \right). \] This simplifies to \( \left( \frac{3}{4} \times 16 + \frac{2}{3} \times 8 \right) - \left( \frac{3}{4} \times 1 + \frac{2}{3} \times (-1) \right). \)

Step 6: Compute the Value

Calculate the final result: \( \left( 12 + \frac{16}{3} \right) - \left( \frac{3}{4} + \left(-\frac{2}{3}\right) \right) = \frac{36}{3} + \frac{16}{3} \) for the first part and \( \frac{3}{4} - \frac{2}{3} = \frac{9}{12} - \frac{8}{12} = \frac{1}{12} \) for the second part. The total is \( \frac{52}{3} - \frac{1}{12} \). Converting into a common denominator, \( \frac{208}{12} - \frac{1}{12} = \frac{207}{12} \).

Key Concepts

Parameterization

Parameterization is a crucial concept in calculus, especially when dealing with line integrals. It allows us to express a curve with a single parameter, usually denoted as \( t \). Consider curve parameterization as a way to "trace out" the path of the curve using a single variable.
In this case, we have a curve defined by the equation \( y = x^2 \). To parameterize this curve, we express both \( x \) and \( y \) in terms of \( t \). Here, this is straightforward: set \( x = t \) and then \( y = t^2 \), which traces out the entire curve. The parameter \( t \) ranges from -1 to 2, allowing us to describe the curve segment from the point (-1,1) to (2,4).
By making this conversion, we can replace \( dx \) and \( dy \) with derivatives \( dt \) during integration, greatly simplifying calculations.

Definite Integral

A definite integral is a fundamental concept in calculus used to calculate the accumulation of quantities, like areas under curves or total values along paths. For line integrals, we extend this idea by integrating over curves or paths, rather than straight intervals.
In a definite integral, we are interested in the overall value between limits \( a \) and \( b \), which are often the start and end points of our parameter, \( t \), in the parameterization. Here, we perform the definite integration over the range from \(-1\) to \(2\) for our parameter \( t \). Our expression is \( \int_{-1}^{2} (3t^3 + 2t^2) \, dt \).
To evaluate, we focus on finding the antiderivatives of the individual terms. After finding the antiderivative, we substitute the limits \( -1 \) and \( 2 \) into the antiderivative to compute the integral's exact value.

Antiderivative

An antiderivative is the inverse operation of finding a derivative. Essentially, when we calculate an antiderivative, we are looking for a function whose derivative matches the function we're interested in. This process is key in evaluating definite integrals.
When tackling the line integral \( \int_{-1}^{2} (3t^3 + 2t^2) \, dt \), we separately find antiderivatives for \( 3t^3 \) and \( 2t^2 \). The antiderivative of \( 3t^3 \) is \( \frac{3}{4}t^4 \), and for \( 2t^2 \), it's \( \frac{2}{3}t^3 \).
These expressions form our combined antiderivative, \( \frac{3}{4}t^4 + \frac{2}{3}t^3 \). By applying the limits of -1 and 2 to this expression, we calculate the net effect of our line integral along the curve. Finding and applying the antiderivative is what allows us to move from a function of change (derivative) back to a function of size or shape.