Pregunta
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29. \( \int_{-1}^{0} e^{x} d x \) 30. \( \int_{-1}^{0} e^{-x} d \) \( \begin{array}{ll}\text { 33. } \int_{0}^{1} \frac{1}{1+x^{2}} d x & \text { 32. } \int_{e}^{1} \frac{1}{x} d x \\ d t] & \text { 34. } \int_{0}^{\sqrt{2} / 2} \frac{1}{\sqrt{1-x^{2}}} d x \\ \text { 35. } \int_{-1}^{8} x^{2 / 3} d x & \text { 36. } \int_{0}^{4} x^{3 / 2} d x\end{array} \)

Ask by Allan Moreno. in the United States
Feb 18,2025

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29. \( 1 - \frac{1}{e} \) 30. \( e - 1 \) 33. \( \frac{\pi}{4} \) 32. \( -1 \) 34. \( \frac{\pi}{4} \) 35. \( \frac{99}{5} \) 36. \( \frac{64}{5} \)

Solución

We will evaluate each of the following integrals one by one. ──────────────────── 29. Evaluate ∫₋₁⁰ eˣ dx The antiderivative of eˣ is eˣ. Thus,   ∫₋₁⁰ eˣ dx = [eˣ]₋₁⁰ = e⁰ – e^(–1) = 1 – 1/e. ──────────────────── 30. Evaluate ∫₋₁⁰ e^(–x) dx The antiderivative of e^(–x) is –e^(–x) (since d/dx[–e^(–x)] = e^(–x)). Thus,   ∫₋₁⁰ e^(–x) dx = [–e^(–x)]₋₁⁰ = [–e⁰] – [–e^(1)] = (–1) – (–e) = e – 1. ──────────────────── 33. Evaluate ∫₀¹ 1/(1+x²) dx The antiderivative of 1/(1+x²) is arctan(x). Thus,   ∫₀¹ 1/(1+x²) dx = [arctan(x)]₀¹ = arctan(1) – arctan(0) = (π/4) – 0 = π/4. ──────────────────── 32. Evaluate ∫ₑ¹ 1/x dx The antiderivative of 1/x is ln|x|. Thus,   ∫ₑ¹ 1/x dx = [ln|x|]ₑ¹ = ln|1| – ln|e| = 0 – 1 = –1. Note: The integration limits are from e to 1 (upper limit smaller than the lower limit), which gives a negative result. ──────────────────── 34. Evaluate ∫₀^(√2⁄2) 1/√(1–x²) dx The antiderivative of 1/√(1–x²) is arcsin(x). Thus,   ∫₀^(√2⁄2) 1/√(1–x²) dx = [arcsin(x)]₀^(√2⁄2) = arcsin(√2⁄2) – arcsin(0) = (π/4) – 0 = π/4. ──────────────────── 35. Evaluate ∫₋₁⁸ x^(2/3) dx First, find the antiderivative:   ∫ x^(2/3) dx = (x^(5/3))/(5/3) + C = (3/5) x^(5/3) + C. Thus,   ∫₋₁⁸ x^(2/3) dx = [(3/5) x^(5/3)]₋₁⁸ = (3/5)[8^(5/3) – (–1)^(5/3)]. Analyze each term:  • 8^(5/3) = (cube root of 8)⁵ = 2⁵ = 32.  • (–1)^(5/3) = (cube root of –1)⁵ = (–1)⁵ = –1. Therefore,   = (3/5)[32 – (–1)] = (3/5)[33] = 99/5. ──────────────────── 36. Evaluate ∫₀⁴ x^(3/2) dx Find the antiderivative:   ∫ x^(3/2) dx = (x^(5/2))/(5/2) + C = (2/5)x^(5/2) + C. Thus,   ∫₀⁴ x^(3/2) dx = [(2/5)x^(5/2)]₀⁴ = (2/5)[4^(5/2) – 0]. Compute 4^(5/2):   4^(5/2) = (√4)⁵ = 2⁵ = 32. So,   = (2/5)*32 = 64/5. ──────────────────── Summary of Answers: 29. 1 – 1/e 30. e – 1 33. π/4 32. –1 34. π/4 35. 99/5 36. 64/5

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Beyond the Answer

To really grasp the magic of integrals, let’s talk about a fun historical fact: The concept of integration dates back to ancient times! The mathematician Archimedes was one of the first to use shapes and areas, essentially laying the groundwork for integral calculus. His innovative methods to compute areas under curves set the stage for what would later evolve into modern calculus. Talk about a brainy legacy! Now, how do we use these integrals in the real world? Imagine you're a civil engineer designing a curvy bridge. Calculating the area under curves helps determine the amount of materials needed! Whether it’s optimizing the design for safety or ensuring aesthetic appeal, integrals play a crucial role in real-world applications like construction, physics, and economics. Isn’t it cool how these mathematical concepts help shape our world?

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