Math Problem Statement

Let G left parenthesis x right parenthesis space equals space integral subscript h subscript 1 left parenthesis x right parenthesis end subscript superscript h subscript 2 left parenthesis x right parenthesis end superscript f left parenthesis t right parenthesis d t where bold italic h subscript bold 1 bold left parenthesis bold italic x bold right parenthesis bold space bold equals bold space bold pi bold space bold italic c bold italic o bold italic s bold left parenthesis bold space bold 2 bold pi bold space bold x bold right parenthesis amd bold italic h subscript bold 2 bold left parenthesis bold italic x bold right parenthesis bold space bold equals bold space bold 4 bold space bold italic t bold italic a bold italic n to the power of bold minus bold 1 end exponent bold left parenthesis bold italic x bold right parenthesis

and f left parenthesis t right parenthesis space equals space199 divided by left parenthesis space 1 space plus space t squared space right parenthesis then Find bold italic G bold left parenthesis bold 1 bold right parenthesis bold space bold plus bold space bold italic G bold apostrophe bold left parenthesis bold 1 bold right parenthesis bold space bold equals

Solution

To solve this problem, we are given the following:

G(x)=∫h1(x)h2(x)f(t) dtG(x) = \int_{h_1(x)}^{h_2(x)} f(t) \, dt

where:

  • h1(x)=πcos⁡(2πx)h_1(x) = \pi \cos(2\pi x)
  • h2(x)=4tan⁡−1(x)h_2(x) = 4 \tan^{-1}(x)
  • f(t)=1991+t2f(t) = \frac{199}{1 + t^2}

We are tasked with finding G(1)+G′(1)G(1) + G'(1).

Step 1: Calculate G(1)G(1)

We need to first evaluate G(1)G(1), which involves calculating the definite integral: G(1)=∫h1(1)h2(1)f(t) dtG(1) = \int_{h_1(1)}^{h_2(1)} f(t) \, dt

  • h1(1)=πcos⁡(2π×1)=πcos⁡(2π)=π×1=πh_1(1) = \pi \cos(2\pi \times 1) = \pi \cos(2\pi) = \pi \times 1 = \pi
  • h2(1)=4tan⁡−1(1)=4×π4=πh_2(1) = 4 \tan^{-1}(1) = 4 \times \frac{\pi}{4} = \pi

Thus, G(1)=∫ππ1991+t2 dtG(1) = \int_{\pi}^{\pi} \frac{199}{1 + t^2} \, dt Since the upper and lower limits of the integral are the same, the value of the integral is zero: G(1)=0G(1) = 0

Step 2: Find G′(x)G'(x)

To find G′(x)G'(x), we use the Leibniz rule for differentiating an integral with variable limits. The formula is: G′(x)=f(h2(x))⋅h2′(x)−f(h1(x))⋅h1′(x)G'(x) = f(h_2(x)) \cdot h_2'(x) - f(h_1(x)) \cdot h_1'(x)

Now, let's compute the derivatives and evaluate them at x=1x = 1:

Derivatives of h1(x)h_1(x) and h2(x)h_2(x)

  • h1(x)=πcos⁡(2πx)h_1(x) = \pi \cos(2\pi x) h1′(x)=−2π2sin⁡(2πx)h_1'(x) = -2\pi^2 \sin(2\pi x) At x=1x = 1, h1′(1)=−2π2sin⁡(2π)=0h_1'(1) = -2\pi^2 \sin(2\pi) = 0.

  • h2(x)=4tan⁡−1(x)h_2(x) = 4 \tan^{-1}(x) h2′(x)=41+x2h_2'(x) = \frac{4}{1 + x^2} At x=1x = 1, h2′(1)=41+12=2h_2'(1) = \frac{4}{1 + 1^2} = 2.

Values of f(h1(1))f(h_1(1)) and f(h2(1))f(h_2(1))

  • h1(1)=πh_1(1) = \pi, so f(h1(1))=f(π)=1991+π2f(h_1(1)) = f(\pi) = \frac{199}{1 + \pi^2}

  • h2(1)=πh_2(1) = \pi, so f(h2(1))=f(π)=1991+π2f(h_2(1)) = f(\pi) = \frac{199}{1 + \pi^2}

Step 3: Compute G′(1)G'(1)

Now using the Leibniz rule: G′(1)=f(h2(1))⋅h2′(1)−f(h1(1))⋅h1′(1)G'(1) = f(h_2(1)) \cdot h_2'(1) - f(h_1(1)) \cdot h_1'(1) Since h1′(1)=0h_1'(1) = 0, the second term vanishes, and we are left with: G′(1)=1991+π2⋅2G'(1) = \frac{199}{1 + \pi^2} \cdot 2 G′(1)=3981+π2G'(1) = \frac{398}{1 + \pi^2}

Step 4: Final Answer

We now add G(1)G(1) and G′(1)G'(1): G(1)+G′(1)=0+3981+π2G(1) + G'(1) = 0 + \frac{398}{1 + \pi^2} Thus, the final result is: 3981+π2\boxed{\frac{398}{1 + \pi^2}}

Would you like more details or have any questions?

Here are five related questions:

  1. How do you compute the derivative of an integral with variable limits?
  2. Why does the value of the integral vanish when the limits are equal?
  3. What is the physical interpretation of Leibniz's rule in calculus?
  4. Can we apply this process to integrals where the limits depend on more complex functions?
  5. How would the answer change if the function f(t)f(t) were different?

Tip: When applying Leibniz's rule, always check if the lower or upper limit's derivative might make a term vanish, simplifying the calculation.

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Leibniz Rule
Trigonometry
Inverse Functions
Differentiation

Formulas

G(x) = ∫ from h1(x) to h2(x) f(t) dt
f(t) = 199 / (1 + t²)
h1(x) = π cos(2πx)
h2(x) = 4 tan⁻¹(x)
Leibniz Rule: d/dx ∫ from a(x) to b(x) f(t) dt = f(b(x))b'(x) - f(a(x))a'(x)

Theorems

Leibniz Rule for Differentiation of Integrals
Fundamental Theorem of Calculus

Suitable Grade Level

College Calculus (Calculus I and II)

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