How do you use the trapezoidal rule to find the integral from 1 to 4 for #6sqrt(lnx)# with n=6?
# int_(1)^(4) 6sqrt(lnx)dx ~~ 15.54800 " "(5dp)#
The values of
Using the trapezoidal rule:
# int_a^bydx ~~ h/2{(y_0+y_n)+2(y_1+y_2+...+y_(n-1))}#
We have:
# int_(1)^(4) 6sqrt(lnx)dx ~~ 0.5/2 { 0 + 7.06446 + 2(3.82057 + 4.99533 + 5.74338 + 6.28888 + 6.71561)} #
# " " = 0.25 { 7.06446 + 2( 27.56378 )}#
# " " = 0.25 { 7.06446 + 55.12755 }#
# " " = 0.25 { 62.19201 }#
# " " = 15.54800#
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To use the trapezoidal rule to approximate the integral (\int_{1}^{4} 6\sqrt{\ln(x)} ,dx) with (n=6) subintervals, follow these steps:
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Determine the width of each subinterval ((h)): The interval ([1, 4]) has a width of (4 - 1 = 3). With (n=6) subintervals, (h = \frac{4 - 1}{6} = \frac{1}{2}).
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Calculate the (x) values (the endpoints of the subintervals): Starting from (x_0 = 1), we increment by (h = \frac{1}{2}) for each subsequent (x). So, the (x) values are (1, 1.5, 2, 2.5, 3, 3.5, 4).
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Evaluate the function at each (x) value: Calculate (6\sqrt{\ln(x_i)}) for each (x_i).
- (f(1) = 6\sqrt{\ln(1)} = 0)
- (f(1.5) = 6\sqrt{\ln(1.5)} \approx 4.477)
- (f(2) = 6\sqrt{\ln(2)} \approx 5.887)
- (f(2.5) = 6\sqrt{\ln(2.5)} \approx 6.952)
- (f(3) = 6\sqrt{\ln(3)} \approx 7.795)
- (f(3.5) = 6\sqrt{\ln(3.5)} \approx 8.491)
- (f(4) = 6\sqrt{\ln(4)} \approx 9.093)
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Apply the trapezoidal rule: The formula for the trapezoidal rule is [ \int_{a}^{b} f(x),dx \approx \frac{h}{2}[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)] ] Substituting the values we've found: [ \int_{1}^{4} 6\sqrt{\ln(x)},dx \approx \frac{1}{2}\left[0 + 2(4.477) + 2(5.887) + 2(6.952) + 2(7.795) + 2(8.491) + 9.093\right] ] Simplifying: [ \approx \frac{1}{2}[0 + 8.954 + 11.774 + 13.904 + 15.59 + 16.982 + 9.093] ] [ = \frac{1}{2} \times 76.297 = 38.1485 ]
Therefore, using the trapezoidal rule with (n=6), the approximation of the integral (\int_{1}^{4} 6\sqrt{\ln(x)} ,dx) is approximately (38.1485).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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