Math Problem Statement

Evaluate the following antiderivatives and integrals.

Solution

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Problem 1:

Part (a):

We are asked to evaluate the integral: (8cos(x)+sec2(x))dx\int \big(8\cos(x) + \sec^2(x)\big) \, dx

  • For 8cos(x)dx\int 8\cos(x) \, dx, we have: 8cos(x)dx=8sin(x)\int 8\cos(x) \, dx = 8\sin(x)

  • For sec2(x)dx\int \sec^2(x) \, dx, we use the standard result: sec2(x)dx=tan(x)\int \sec^2(x) \, dx = \tan(x)

Adding these, we get: (8cos(x)+sec2(x))dx=8sin(x)+tan(x)+C\int \big(8\cos(x) + \sec^2(x)\big) \, dx = 8\sin(x) + \tan(x) + C

Part (b):

We are asked to evaluate the integral: (x6+3xx4)dx\int \left(x^6 + \frac{3\sqrt{x}}{x^4}\right) \, dx

Simplify the second term: 3xx4=3x3.5\frac{3\sqrt{x}}{x^4} = 3x^{-3.5}

Thus, the integral becomes: (x6+3x3.5)dx\int \big(x^6 + 3x^{-3.5}\big) \, dx

  • For x6dx\int x^6 \, dx, use the power rule: x6dx=x77\int x^6 \, dx = \frac{x^7}{7}

  • For 3x3.5dx\int 3x^{-3.5} \, dx, use the power rule: 3x3.5dx=3x2.52.5=65x2.5\int 3x^{-3.5} \, dx = 3 \cdot \frac{x^{-2.5}}{-2.5} = -\frac{6}{5}x^{-2.5}

Combining these: (x6+3xx4)dx=x7765x2.5+C\int \left(x^6 + \frac{3\sqrt{x}}{x^4}\right) \, dx = \frac{x^7}{7} - \frac{6}{5}x^{-2.5} + C


Problem 2:

We are asked to evaluate: cos(7x)dx\int \cos(7x) \, dx

Let u=7xu = 7x, so du=7dxdu = 7 \, dx or dx=du7dx = \frac{du}{7}. Substitute into the integral: cos(7x)dx=17cos(u)du\int \cos(7x) \, dx = \frac{1}{7} \int \cos(u) \, du

The integral of cos(u)\cos(u) is sin(u)\sin(u), so: 17cos(u)du=17sin(u)\frac{1}{7} \int \cos(u) \, du = \frac{1}{7} \sin(u)

Substitute back u=7xu = 7x: cos(7x)dx=17sin(7x)+C\int \cos(7x) \, dx = \frac{1}{7} \sin(7x) + C


Problem 3:

We are tasked to find f(x)f(x) such that: f(x)=x26xandf(1)=2f'(x) = x^2 - 6x \quad \text{and} \quad f(1) = 2

Integrate f(x)f'(x): f(x)=(x26x)dxf(x) = \int (x^2 - 6x) \, dx

  • For x2dx\int x^2 \, dx: x2dx=x33\int x^2 \, dx = \frac{x^3}{3}

  • For 6xdx\int -6x \, dx: 6xdx=3x2\int -6x \, dx = -3x^2

Combine: f(x)=x333x2+Cf(x) = \frac{x^3}{3} - 3x^2 + C

To find CC, use the condition f(1)=2f(1) = 2: f(1)=1333(12)+C=2f(1) = \frac{1^3}{3} - 3(1^2) + C = 2

Simplify: 133+C=2\frac{1}{3} - 3 + C = 2

C=2+313=513=15313=143C = 2 + 3 - \frac{1}{3} = 5 - \frac{1}{3} = \frac{15}{3} - \frac{1}{3} = \frac{14}{3}

Thus: f(x)=x333x2+143f(x) = \frac{x^3}{3} - 3x^2 + \frac{14}{3}


If you'd like further clarification or detailed steps, let me know! Here are 5 related questions to explore further:

  1. How do you verify the correctness of an antiderivative?
  2. What are the rules for integrating functions involving trigonometric terms?
  3. How do substitution methods simplify certain integrals?
  4. What is the significance of initial conditions in determining constants of integration?
  5. How do power rules change for negative or fractional exponents?

Tip: Always double-check your work by differentiating your final answer to ensure it matches the original integrand!

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Math Problem Analysis

Mathematical Concepts

Antiderivatives
Integration
Power Rule
Trigonometric Integrals
Initial Conditions

Formulas

∫cos(x) dx = sin(x) + C
∫sec^2(x) dx = tan(x) + C
∫x^n dx = x^(n+1) / (n+1) + C for n ≠ -1
Substitution Method

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

College Calculus or Advanced High School Calculus