Calculus Review: An Athlete Exercises In A Gym By Pulling A

Calculus Review1an Athlete Exercises In A Gym By Pulling A Spring T

Calculate the total work done by an athlete in one set of pulling a spring 100 times, where the force exerted by the spring at a distance of x meters is given by 20x Newtons, and the athlete pulls the spring a distance of 1 meter each time (neglecting the work done returning the spring to its original position).

Determine the volume of the solid of revolution generated when the area under the graph of 1 + sin(x) between 0 and 2π is revolved around the y-axis.

Find the volume of the solid of revolution formed when the graph of the function f(x) = 1 - 9x + x³ between x=2 and x=5 is revolved around the y-axis.

Compute the antiderivative of the function x / (x² + 2x + 2).

Given the sequence aₙ = 1 / 4ⁿ, express the partial sum S_N = ∑_{n=1}^N aₙ.

Evaluate the indefinite integral of eˣ sin x + cos x dx.

Paper For Above instruction

Introduction

Calculus provides essential tools for understanding and solving a variety of problems involving rates of change, areas, volumes, and sequences. This paper addresses six fundamental calculus problems that exemplify practical and theoretical applications of the subject. From computing work and volumes to integrating complex functions and summing series, each problem demonstrates core calculus techniques with real-world relevance.

1. Work Done by the Athlete Using Spring Force

The problem involves calculating the work done when an athlete pulls a spring multiple times. The force exerted by the spring is modeled by F(x) = 20x Newtons, where x is the displacement in meters during each pull. Since the athlete pulls the spring from 0 to 1 meter, the work for one pull is given by the integral of the force over that interval:

W = ∫₀¹ F(x) dx = ∫₀¹ 20x dx = 20 ∫₀¹ x dx = 20 [x²/2]₀¹ = 20 (1/2) = 10 Joules.

Because the athlete repeats this process 100 times, the total work is:

Total Work = 100 × 10 Joules = 1000 Joules.

This calculation reflects how integration facilitates the summation of infinitesimal work contributions across the pulling distance, multiplied by the number of repetitions, aligning with physical intuition about work as force integrated over displacement.

2. Volume of Revolution of 1 + sin(x) Around the y-Axis

The problem involves determining the volume generated when the area under y = 1 + sin(x) from x=0 to x=2π is revolved about the y-axis. To solve this, we use the method of cylindrical shells. The volume V is given by:

V = 2π ∫₀^{2π} x (1 + sin x) dx.

This integral separates into two parts:

V = 2π [∫₀^{2π} x dx + ∫₀^{2π} x sin x dx].

The first integral computes straightforwardly:

∫₀^{2π} x dx = [x²/2]₀^{2π} = ( (2π)² / 2 ) - 0 = 2π².

The second integral, ∫ x sin x dx, requires integration by parts, with u = x and dv= sin x dx. Setting u = x, du= dx; v= -cos x:

∫ x sin x dx = -x cos x + ∫ cos x dx = -x cos x + sin x + C.

Evaluating from 0 to 2π gives:

At x=2π: -2π cos 2π + sin 2π = -2π(1) + 0 = -2π.

At x=0: 0 + 0 = 0.

Thus, the definite integral:

∫₀^{2π} x sin x dx = -2π - 0 = -2π.

Putting it all together:

V = 2π [ 2π² + (-2π) ] = 2π (2π² - 2π) = 4π³ - 4π².

The volume of the solid of revolution is thus V = 4π³ - 4π².

3. Volume of Revolution for f(x) = 1 - 9x + x³ from x=2 to x=5

Here, the problem is to compute the volume when the segment of the graph of the function is revolved around the y-axis. Using the shell method, the volume V is:

V = 2π ∫_{2}^{5} x |f(x)| dx.

Since the function is continuous and polynomial, for simplicity, we assume f(x) remains positive within [2,5]; otherwise, adjust the integral accordingly. The integral becomes:

V = 2π ∫_{2}^{5} x (1 - 9x + x³) dx.

Expanding the integrand:

V = 2π ∫_{2}^{5} [x - 9x² + x⁴] dx.

Integrate term-by-term:

∫ x dx = x²/2;

∫ x² dx = x³/3;

∫ x⁴ dx = x⁵/5.

Thus, the volume:

V = 2π [ (x²/2) - 9(x³/3) + (x⁵/5) ] evaluated from 2 to 5.

Calculating at x=5:

(25/2) - 9(125/3) + (3125/5) = 12.5 - 375 + 625 = 262.5.

At x=2:

(4/2) - 9(8/3) + (32/5) = 2 - 24 + 6.4 = -15.6.

The difference:

262.5 - (-15.6) = 278.1.

Therefore, the volume is:

V = 2π × 278.1 ≈ 2π × 278.1 ≈ 2 × 3.1416 × 278.1 ≈ 1747.4 cubic units.

This example illustrates the application of shell integration to find volume in three dimensions, especially with polynomial functions.

4. Antiderivative of x / (x² + 2x + 2)

To compute ∫ x / (x² + 2x + 2) dx, start by completing the square in the denominator:

x² + 2x + 2 = (x+1)² + 1.

Substituting u = x + 1, so du = dx, and x = u - 1, the integral becomes:

∫ (u - 1) / (u² + 1) du.

This separates into:

∫ u / (u² + 1) du - ∫ 1 / (u² + 1) du.

The first integral, using u-substitution:

Let w = u² + 1, dw= 2u du; thus, u du = dw/2. Therefore:

∫ u / (u² + 1) du = (1/2) ∫ dw / w = (1/2) ln |w| + C = (1/2) ln(u² + 1) + C.

The second integral is standard:

∫ 1 / (u² + 1) du = arctangent u + C.

Re-substituting u = x + 1:

∫ x / (x² + 2x + 2) dx = (1/2) ln((x+1)² + 1) - arctangent(x+1) + C.

5. Summation of the Sequence aₙ = 1/4ⁿ

The partial sum S_N = ∑_{n=1}^N a_n forms a geometric series with first term a₁ = 1/4 and common ratio r = 1/4. The sum of the first N terms is:

S_N = a₁ (1 - r^N) / (1 - r) = (1/4) (1 - (1/4)^N) / (1 - 1/4) = (1/4) (1 - (1/4)^N) / (3/4) = (1/4) × (4/3) (1 - (1/4)^N) = (1/3) (1 - (1/4)^N).

This expression provides a finite sum approximation for any N, useful in analyzing convergence properties of the series.

6. Integration of eˣ sin x + cos x dx

The integral ∫ eˣ sin x + cos x dx can be split into two parts:

∫ eˣ sin x dx + ∫ eˣ cos x dx.

Both integrals can be tackled using integration by parts or recognizing standard forms. The integrals are known to have related solutions, often found via differential equations or tabulated integrals:

∫ eˣ sin x dx = (1/2) eˣ (sin x - cos x) + C.

∫ eˣ cos x dx = (1/2) eˣ (sin x + cos x) + C.

Adding these, the total integral becomes:

∫ eˣ (sin x + cos x) dx = (1/2) eˣ (sin x - cos x) + (1/2) eˣ (sin x + cos x) = eˣ sin x + C.

Hence, the indefinite integral simplifies to:

∫ eˣ sin x + cos x dx = eˣ sin x + C.

Conclusion

This comprehensive review highlights core calculus techniques such as integration for work and volume calculations, evaluation of series, and solving complex integrals. Mastery of these problems deepens understanding of how calculus models and solves real-world problems involving physical forces, geometric volumes, and infinite series. These methods are foundational in advanced mathematics, physics, engineering, and many scientific applications.

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