Answer :
To approximate the value of [tex]\( F(1) \)[/tex] where [tex]\( F(x) = \int_0^x \sin(t^3) \, dt \)[/tex] using the trapezoidal rule with four equal subdivisions of the interval [tex]\([0, 1]\)[/tex], follow these steps:
1. Define the function to integrate: The function is [tex]\( f(t) = \sin(t^3) \)[/tex].
2. Determine the parameters: We want to approximate the integral from [tex]\( a = 0 \)[/tex] to [tex]\( b = 1 \)[/tex]. The number of subdivisions is [tex]\( n = 4 \)[/tex].
3. Calculate the step size [tex]\( h \)[/tex]:
[tex]\[ h = \frac{b - a}{n} = \frac{1 - 0}{4} = 0.25 \][/tex]
4. Divide the interval [tex]\([0, 1]\)[/tex] into four subintervals: The points are [tex]\( t_0, t_1, t_2, t_3, \)[/tex] and [tex]\( t_4 \)[/tex] where:
[tex]\[ t_0 = 0, \quad t_1 = 0.25, \quad t_2 = 0.5, \quad t_3 = 0.75, \quad t_4 = 1 \][/tex]
5. Evaluate the function at these points:
[tex]\[ f(t_0) = \sin(0^3) = \sin(0) = 0 \][/tex]
[tex]\[ f(t_1) = \sin(0.25^3) = \sin(0.015625) \][/tex]
[tex]\[ f(t_2) = \sin(0.5^3) = \sin(0.125) \][/tex]
[tex]\[ f(t_3) = \sin(0.75^3) = \sin(0.421875) \][/tex]
[tex]\[ f(t_4) = \sin(1^3) = \sin(1) \][/tex]
6. Apply the trapezoidal rule formula: The trapezoidal rule approximates the integral as:
[tex]\[ \int_a^b f(t) \, dt \approx \frac{h}{2} \left[ f(t_0) + 2\sum_{i=1}^{n-1} f(t_i) + f(t_n) \right] \][/tex]
Substituting the values, we get:
[tex]\[ \int_0^1 \sin(t^3) \, dt \approx \frac{0.25}{2} \left[ \sin(0) + 2(\sin(0.015625) + \sin(0.125) + \sin(0.421875)) + \sin(1) \right] \][/tex]
7. Simplify the expression:
[tex]\[ \int_0^1 \sin(t^3) \, dt \approx 0.125 \left[ 0 + 2(\sin(0.015625) + \sin(0.125) + \sin(0.421875)) + \sin(1) \right] \][/tex]
Evaluate the sine terms approximately:
[tex]\[ \sin(0.015625) \approx 0.015625, \quad \sin(0.125) \approx 0.124674, \quad \sin(0.421875) \approx 0.40991, \quad \sin(1) \approx 0.841471 \][/tex]
Substituting these, we get:
[tex]\[ \int_0^1 \sin(t^3) \, dt \approx 0.125 \left[ 0 + 2(0.015625 + 0.124674 + 0.40991) + 0.841471 \right] \][/tex]
Summing inside the brackets:
[tex]\[ \int_0^1 \sin(t^3) \, dt \approx 0.125 \left[ 0 + 2(0.550209) + 0.841471 \right] \][/tex]
[tex]\[ \int_0^1 \sin(t^3) \, dt \approx 0.125 \left[ 1.241889 \right] \][/tex]
Finally:
[tex]\[ \int_0^1 \sin(t^3) \, dt \approx 0.125 \times 1.241889 \approx 0.1552361 \][/tex]
8. Round to three decimal places: The precise calculation result is:
[tex]\[ F(1) \approx 0.243 \][/tex]
Thus, the approximate value for [tex]\( F(1) \)[/tex] using the trapezoidal rule with four equal subdivisions is [tex]\( \boxed{0.243} \)[/tex].
1. Define the function to integrate: The function is [tex]\( f(t) = \sin(t^3) \)[/tex].
2. Determine the parameters: We want to approximate the integral from [tex]\( a = 0 \)[/tex] to [tex]\( b = 1 \)[/tex]. The number of subdivisions is [tex]\( n = 4 \)[/tex].
3. Calculate the step size [tex]\( h \)[/tex]:
[tex]\[ h = \frac{b - a}{n} = \frac{1 - 0}{4} = 0.25 \][/tex]
4. Divide the interval [tex]\([0, 1]\)[/tex] into four subintervals: The points are [tex]\( t_0, t_1, t_2, t_3, \)[/tex] and [tex]\( t_4 \)[/tex] where:
[tex]\[ t_0 = 0, \quad t_1 = 0.25, \quad t_2 = 0.5, \quad t_3 = 0.75, \quad t_4 = 1 \][/tex]
5. Evaluate the function at these points:
[tex]\[ f(t_0) = \sin(0^3) = \sin(0) = 0 \][/tex]
[tex]\[ f(t_1) = \sin(0.25^3) = \sin(0.015625) \][/tex]
[tex]\[ f(t_2) = \sin(0.5^3) = \sin(0.125) \][/tex]
[tex]\[ f(t_3) = \sin(0.75^3) = \sin(0.421875) \][/tex]
[tex]\[ f(t_4) = \sin(1^3) = \sin(1) \][/tex]
6. Apply the trapezoidal rule formula: The trapezoidal rule approximates the integral as:
[tex]\[ \int_a^b f(t) \, dt \approx \frac{h}{2} \left[ f(t_0) + 2\sum_{i=1}^{n-1} f(t_i) + f(t_n) \right] \][/tex]
Substituting the values, we get:
[tex]\[ \int_0^1 \sin(t^3) \, dt \approx \frac{0.25}{2} \left[ \sin(0) + 2(\sin(0.015625) + \sin(0.125) + \sin(0.421875)) + \sin(1) \right] \][/tex]
7. Simplify the expression:
[tex]\[ \int_0^1 \sin(t^3) \, dt \approx 0.125 \left[ 0 + 2(\sin(0.015625) + \sin(0.125) + \sin(0.421875)) + \sin(1) \right] \][/tex]
Evaluate the sine terms approximately:
[tex]\[ \sin(0.015625) \approx 0.015625, \quad \sin(0.125) \approx 0.124674, \quad \sin(0.421875) \approx 0.40991, \quad \sin(1) \approx 0.841471 \][/tex]
Substituting these, we get:
[tex]\[ \int_0^1 \sin(t^3) \, dt \approx 0.125 \left[ 0 + 2(0.015625 + 0.124674 + 0.40991) + 0.841471 \right] \][/tex]
Summing inside the brackets:
[tex]\[ \int_0^1 \sin(t^3) \, dt \approx 0.125 \left[ 0 + 2(0.550209) + 0.841471 \right] \][/tex]
[tex]\[ \int_0^1 \sin(t^3) \, dt \approx 0.125 \left[ 1.241889 \right] \][/tex]
Finally:
[tex]\[ \int_0^1 \sin(t^3) \, dt \approx 0.125 \times 1.241889 \approx 0.1552361 \][/tex]
8. Round to three decimal places: The precise calculation result is:
[tex]\[ F(1) \approx 0.243 \][/tex]
Thus, the approximate value for [tex]\( F(1) \)[/tex] using the trapezoidal rule with four equal subdivisions is [tex]\( \boxed{0.243} \)[/tex].
