91影视

For the first pair of integrals, we notice that the functions \(2^{x^2}\) and \(2^{x^3}\) are both increasing over the interval [0, 1] because their exponentials are increasing (as \(x^2 \leq x^3\) for \(x\) in [0, 1]). Thus, we can say that \(2^{x^2} \leq 2^{x^3}\) for \(x\in [0, 1]\). Since a smaller integrand results in a smaller integral, it follows that \(\int_{0}^{1} 2^{x^{2}} d x \leq \int_{0}^{1} 2^{x^{3}} d x\). #setup(expression)# x = symbols('x') #setup_tool(latex_code)# x_square = 2**(x**2) #setup_tool(latex_code)# x_cube = 2**(x**3) #setup_tool(latex_code)# integral_x_square = integrate(x_square, (x, 0, 1)) #setup_tool(latex_code)# integral_x_cube = integrate(x_cube, (x, 0, 1))

For the second pair of integrals, we notice that the functions \(2^{x^2}\) and \(2^{x^3}\) are both increasing over the interval [1, 2]. However, for \(x\) in [1, 2], we have \(x^2 \geq x^3\), which means \(2^{x^2} \geq 2^{x^3}\). Thus, we can say that \(\int_{1}^{2} 2^{x^{2}} d x \geq \int_{1}^{2} 2^{x^{3}} d x\). #setup_tool(latex_code)# integral_x_square_1_2 = integrate(x_square, (x, 1, 2)) #setup_tool(latex_code)# integral_x_cube_1_2 = integrate(x_cube, (x, 1, 2))

For the third pair of integrals, we have the integrands \(\ln x\) and \((\ln x)^2\). Since \(\ln x\) is a monotonically increasing and concave function over the interval [1, 2], we can conclude that the function \((\ln x)^2\) is also monotonically increasing and convex over the interval [1, 2]. Thus, we can say that \(\ln x \leq (\ln x)^2\) and therefore, \(\int_{1}^{2} \ln x d x \leq \int_{1}^{2} (\ln x)^{2} d x\). #setup_tool(latex_code)# ln_x = ln(x) #setup_tool(latex_code)# ln_x_square = ln(x)**2 #setup_tool(latex_code)# integral_ln_x = integrate(ln_x, (x, 1, 2)) #setup_tool(latex_code)# integral_ln_x_square = integrate(ln_x_square, (x, 1, 2))

For the fourth pair of integrals, we again have the integrands \(\ln x\) and \((\ln x)^2\). As we have already mentioned in Step 3, since \(\ln x\) is a monotonically increasing and concave function over the interval [1, 2], the function \((\ln x)^2\) is monotonically increasing and convex over the interval [1, 2]. However, the interval is [3, 4] for this pair. It turns out that these properties still hold as the inequality between \(\ln x\) and \((\ln x)^2\) is maintained. Thus, we can say that \(\ln x \leq (\ln x)^2\) and therefore, \(\int_{3}^{4} \ln x d x \leq \int_{3}^{4} (\ln x)^{2} d x\). #setup_tool(latex_code)# integral_ln_x_3_4 = integrate(ln_x, (x, 3, 4)) #setup_tool(latex_code)# integral_ln_x_square_3_4 = integrate(ln_x_square, (x, 3, 4)) We have successfully compared all the pairs of integrals: (i) \(\int_{0}^{1} 2^{x^{2}} d x\) is less than \(\int_{0}^{1} 2^{x^{3}} d x\). (ii) \(\int_{1}^{2} 2^{x^{2}} d x\) is greater than \(\int_{1}^{2} 2^{x^{3}} d x\). (iii) \(\int_{1}^{2} \ln x d x\) is less than \(\int_{1}^{2} (\ln x)^{2} d x\). (iv) \(\int_{3}^{4} \ln x d x\) is less than \(\int_{3}^{4} (\ln x)^{2} d x\).