horario AHUB

Prévia do material em texto

**Explicação**: A integral é: 
 \[ 
 \int_0^1 (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_0^1 = \left( 1 - \frac{1}{3} \right) = 
\frac{2}{3} 
 \] 
 
37. **Problema 37**: Calcule \( \int_0^{\pi/2} \sin^2(x) \, dx \). 
 a) \( \frac{\pi}{4} \) 
 b) \( \frac{\pi}{6} \) 
 c) \( \frac{\pi}{3} \) 
 d) \( \frac{\pi}{2} \) 
 **Resposta**: a) \( \frac{\pi}{4} \) 
 **Explicação**: Usando a identidade \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \): 
 \[ 
 \int_0^{\pi/2} \sin^2(x) \, dx = \frac{1}{2} \left[ x - \frac{\sin(2x)}{2} \right]_0^{\pi/2} = 
\frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4} 
 \] 
 
38. **Problema 38**: Calcule o limite \( \lim_{x \to 0} \frac{x^2}{\sin(x)} \). 
 a) \( 0 \) 
 b) \( 1 \) 
 c) \( \infty \) 
 d) \( -1 \) 
 **Resposta**: a) \( 0 \) 
 **Explicação**: Usando a regra de L'Hôpital: 
 \[ 
 \lim_{x \to 0} \frac{x^2}{\sin(x)} = \lim_{x \to 0} \frac{2x}{\cos(x)} = 0 
 \] 
 
39. **Problema 39**: Calcule \( \int_0^1 (x^5 - 3x^4 + 3x^3 - x^2) \, dx \). 
 a) \( 0 \) 
 b) \( \frac{1}{6} \) 
 c) \( \frac{1}{4} \) 
 d) \( \frac{1}{3} \) 
 **Resposta**: a) \( 0 \) 
 **Explicação**: A integral é: 
 \[ 
 \int_0^1 (x^5 - 3x^4 + 3x^3 - x^2) \, dx = \left[ \frac{x^6}{6} - \frac{3x^5}{5} + \frac{3x^4}{4} 
- \frac{x^3}{3} \right]_0^1 = \left( \frac{1}{6} - \frac{3}{5} + \frac{3}{4} - \frac{1}{3} \right) = 0 
 \] 
 
40. **Problema 40**: Calcule \( \int_0^1 (2x^2 - 4x + 2) \, dx \). 
 a) \( 0 \) 
 b) \( 1 \) 
 c) \( \frac{1}{3} \) 
 d) \( 2 \) 
 **Resposta**: b) \( 1 \) 
 **Explicação**: A integral é: 
 \[ 
 \int_0^1 (2x^2 - 4x + 2) \, dx = \left[ \frac{2x^3}{3} - 2x^2 + 2x \right]_0^1 = \left( \frac{2}{3} 
- 2 + 2 \right) = \frac{2}{3} 
 \] 
 
41. **Problema 41**: Calcule \( \int \frac{1}{1+x^2} \, dx \). 
 a) \( \tan^{-1}(x) + C \) 
 b) \( \frac{1}{1+x} + C \) 
 c) \( \ln(1+x^2) + C \) 
 d) \( \frac{1}{2} \tan^{-1}(x) + C \) 
 **Resposta**: a) \( \tan^{-1}(x) + C \) 
 **Explicação**: A integral é: 
 \[ 
 \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) + C 
 \] 
 
42. **Problema 42**: Calcule \( \lim_{x \to 0} \frac{\ln(1+x)}{x} \). 
 a) \( 0 \) 
 b) \( 1 \) 
 c) \( \infty \) 
 d) \( -1 \) 
 **Resposta**: b) \( 1 \) 
 **Explicação**: Usando a regra de L'Hôpital: 
 \[ 
 \lim_{x \to 0} \frac{\ln(1+x)}{x} = \lim_{x \to 0} \frac{\frac{1}{1+x}}{1} = 1 
 \] 
 
43. **Problema 43**: Calcule \( \int_0^1 (3x^2 - 4x + 1) \, dx \). 
 a) \( 0 \) 
 b) \( \frac{1}{3} \) 
 c) \( \frac{1}{6} \) 
 d) \( 1 \) 
 **Resposta**: a) \( 0 \) 
 **Explicação**: A integral é: 
 \[ 
 \int_0^1 (3x^2 - 4x + 1) \, dx = \left[ x^3 - 2x^2 + x \right]_0^1 = (1 - 2 + 1) = 0 
 \] 
 
44. **Problema 44**: Calcule \( \int_0^{\pi/2} \sin^3(x) \, dx \). 
 a) \( \frac{3}{8} \) 
 b) \( \frac{1}{4} \) 
 c) \( \frac{1}{2} \) 
 d) \( \frac{1}{3} \) 
 **Resposta**: a) \( \frac{3}{8} \) 
 **Explicação**: Usando a identidade \( \sin^3(x) = \sin(x)(1 - \cos^2(x)) \): 
 \[ 
 \int_0^{\pi/2} \sin^3(x) \, dx = \int_0^{\pi/2} \sin(x) - \sin(x)\cos^2(x) \, dx = \frac{3}{8}