Problem 12 - Basic Technical Mathematics with Calculus

Question

Find the areas bounded by the indicated curves. $$y=3 \sqrt{x+1}, x=0, y=6$$

Step-by-Step Solution

Answer

The area bounded by the given curves is 14 square units.

Step 1: Identify the Area to be Calculated

First, identify the bounded region formed by the curves. The curve given is \(y = 3\sqrt{x+1}\), the vertical line \(x = 0\), and the horizontal line \(y = 6\). The area is bounded by these lines and the curve.

Step 2: Find the Intersection Points

Find the intersection points of the curve \(y = 3\sqrt{x+1}\) with the horizontal line \(y = 6\). Set \(3\sqrt{x+1} = 6\) and solve for \(x\):\[ 3\sqrt{x+1} = 6 \]Divide both sides by 3:\[ \sqrt{x+1} = 2 \]Square both sides:\[ x+1 = 4 \]Subtract 1 to get:\[ x = 3 \]So the intersection point is \((3, 6)\). The region to compute is from \(x=0\) to \(x=3\).

Step 3: Set Up the Integral

Set up the integral of the curve \(y = 3\sqrt{x+1}\) with respect to \(x\) from 0 to 3. The area under the curve and above the x-axis is defined by the integral:\[ \int_{0}^{3} 3\sqrt{x+1} \, dx \]

Step 4: Solve the Integral

To solve the integral \(\int 3\sqrt{x+1} \, dx\), use a substitution method. Let \(u = x + 1\) so that \(du = dx\):\[ \int \, 3\sqrt{x+1} \, dx = \int \, 3 \sqrt{u} \, du = 3\int u^{1/2} \, du \]Integrate using the power rule:\[ 3\int u^{1/2} \, du = 3 \left( \frac{u^{3/2}}{3/2} \right) + C = 2u^{3/2} + C \]Re-substitute \(u = x+1\):\[ 2(x+1)^{3/2} + C \]

Step 5: Evaluate the Definite Integral

Evaluate the definite integral from 0 to 3:\[ \left[ 2(x+1)^{3/2} \right]_0^3 \]Calculate at the bounds:\[ 2(3+1)^{3/2} - 2(0+1)^{3/2} \]\[ = 2(4^{3/2}) - 2(1^{3/2}) \]\[ = 2(8) - 2(1) \]\[ = 16 - 2 \]\[ = 14 \]

Step 6: Conclusion: Area Calculated

The area bounded by the curve \(y = 3\sqrt{x+1}\), the line \(x = 0\), and the line \(y = 6\) is 14 square units.

Key Concepts

Definite Integral

A definite integral is a powerful tool in calculus that helps us find the area under a curve within a specific interval. This means you're not just finding the function's behavior but measuring the space it covers. For example, with the curve \(y = 3\sqrt{x+1}\), the definite integral measures the area under this curve from \(x = 0\) to \(x = 3\). Definite integrals are like summing an infinite number of infinitesimally thin rectangles under the curve. This approach allows us to measure exact areas, which is crucial in understanding the space the curve contains.
The notation \(\int_{a}^{b} f(x) \, dx\) refers to the definite integral of \(f(x)\) from \(x=a\) to \(x=b\). It's essential for solving problems involving area, physics, and engineering by providing precise quantities for areas and sums.

Area Under a Curve

Finding the area under a curve relies heavily on integration. This area represents the total value or space under the plotted curve of a function up to a certain point. In our example, we're interested in the region between \(x = 0\) and \(x = 3\). This involves computing the integral of the function \(y = 3\sqrt{x+1}\), giving us the exact bounded area.

• This concept is vital in various real-world applications, including calculating distances, volumes, and even economic profits.
• By solving the integral of this function, you determine the area under the curve, turning an abstract problem into a tangible measurement, resulting in 14 square units in this instance.

Newcomers might find this approach abstract, but practice makes it easier to visualize and solve these problems.

Substitution Method

The substitution method in integration simplifies computing more challenging integrals by changing variables. In calculus, you often encounter complex functions, and substitution allows you to rewrite these in a simpler form. For the function \(3\sqrt{x+1}\), we let \(u = x + 1\). Here, changing the variable from \(x\) to \(u\) turns a challenging square root into an easier power function. This method lightens the load on complexity, transforming the integral into one involving \(u^{1/2}\).
Once integrated, substituting back to the original variable converts the result into a comprehensive solution for the real-world problem.
The substitution method is invaluable, especially when tackling non-linear equations that don't lend themselves to simple integration techniques.

Intersection Points

Intersection points are where two or more graphs meet, marking crucial changes in inclination, behavior, or enclosed areas. In this exercise, we examine where the curve \(y = 3\sqrt{x+1}\) meets the line \(y = 6\). Solving \(3\sqrt{x+1} = 6\) provides the intersection at \(x = 3\).

• Why is this important? It identifies the bounds of the area under the curve we are examining.
• These points essentially signify where significant shifts occur in graphs or where boundaries for integration are defined.

Finding these points allows us to set limits for our definite integral, ensuring an accurate calculation of the area. This understanding is central to problems like calculating bounded areas, helping dictate the intervals you integrate over.