Math  /  Calculus

QuestionLet f(x)=x2+xf(x)=\sqrt{x^{2}+x} for 1x71 \leq x \leq 7. We wish to estimate 17f(x)dx\int_{1}^{7} f(x) d x by the Trapezoidal Rule. a) Divide the domain of ff into 4 sub-intervals of equal length. Calculate their common length Δx\Delta x (exact value). 1.5 \square b) Find the approximation of 17f(x)dx\int_{1}^{7} f(x) d x that the Trapezoidal Rule produces with 4 sub-intervals. Give the answer with ±0.0001\pm 0.0001 precision. Number

Studdy Solution

STEP 1

1. The function f(x)=x2+x f(x) = \sqrt{x^2 + x} is continuous and defined on the interval [1,7][1, 7].
2. We will use the Trapezoidal Rule to estimate the integral of f(x) f(x) over [1,7][1, 7].
3. The interval [1,7][1, 7] will be divided into 4 equal sub-intervals.

STEP 2

1. Calculate the common length Δx\Delta x of the sub-intervals.
2. Apply the Trapezoidal Rule to estimate the integral.

STEP 3

Calculate the common length Δx\Delta x of the sub-intervals:
The interval [1,7][1, 7] has a total length of 71=67 - 1 = 6.
Divide this interval into 4 equal sub-intervals:
Δx=64=1.5\Delta x = \frac{6}{4} = 1.5

STEP 4

Apply the Trapezoidal Rule to estimate the integral:
The Trapezoidal Rule formula for n n sub-intervals is:
abf(x)dxΔx2(f(x0)+2f(x1)+2f(x2)++2f(xn1)+f(xn))\int_{a}^{b} f(x) \, dx \approx \frac{\Delta x}{2} \left( f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n) \right)
Where x0,x1,,xn x_0, x_1, \ldots, x_n are the endpoints of the sub-intervals.
For our problem: - a=1 a = 1 , b=7 b = 7 , n=4 n = 4 - Δx=1.5\Delta x = 1.5 - The endpoints are x0=1 x_0 = 1 , x1=2.5 x_1 = 2.5 , x2=4 x_2 = 4 , x3=5.5 x_3 = 5.5 , x4=7 x_4 = 7 .
Calculate f(x) f(x) at each endpoint:
f(x0)=f(1)=12+1=2f(x_0) = f(1) = \sqrt{1^2 + 1} = \sqrt{2} f(x1)=f(2.5)=(2.5)2+2.5=6.25+2.5=8.75f(x_1) = f(2.5) = \sqrt{(2.5)^2 + 2.5} = \sqrt{6.25 + 2.5} = \sqrt{8.75} f(x2)=f(4)=42+4=16+4=20f(x_2) = f(4) = \sqrt{4^2 + 4} = \sqrt{16 + 4} = \sqrt{20} f(x3)=f(5.5)=(5.5)2+5.5=30.25+5.5=35.75f(x_3) = f(5.5) = \sqrt{(5.5)^2 + 5.5} = \sqrt{30.25 + 5.5} = \sqrt{35.75} f(x4)=f(7)=72+7=49+7=56f(x_4) = f(7) = \sqrt{7^2 + 7} = \sqrt{49 + 7} = \sqrt{56}
Apply the Trapezoidal Rule:
17f(x)dx1.52(2+28.75+220+235.75+56)\int_{1}^{7} f(x) \, dx \approx \frac{1.5}{2} \left( \sqrt{2} + 2\sqrt{8.75} + 2\sqrt{20} + 2\sqrt{35.75} + \sqrt{56} \right)
=0.75(2+28.75+220+235.75+56)= 0.75 \left( \sqrt{2} + 2\sqrt{8.75} + 2\sqrt{20} + 2\sqrt{35.75} + \sqrt{56} \right)
Calculate the numerical approximation:
0.75(1.4142+2×2.9580+2×4.4721+2×5.9791+7.4833)\approx 0.75 \left( 1.4142 + 2 \times 2.9580 + 2 \times 4.4721 + 2 \times 5.9791 + 7.4833 \right)
0.75(1.4142+5.9160+8.9442+11.9582+7.4833)\approx 0.75 \left( 1.4142 + 5.9160 + 8.9442 + 11.9582 + 7.4833 \right)
0.75×35.7159\approx 0.75 \times 35.7159
26.7869\approx 26.7869
The approximation of 17f(x)dx\int_{1}^{7} f(x) \, dx using the Trapezoidal Rule with 4 sub-intervals is approximately 26.7869\boxed{26.7869}.

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