Question 705056

SCQMEDIUM

The integral $I = \int_0^1 \frac{2x}{1+x^2} dx$ is estimated using Simpson's $1/3$ rd rule with a grid value of $h = 0.5$ . The difference $(I_{estimated} - I_{exact})$ is closest to

(A)

$0.007$

(B)

$0.001$

(C)

$0.0007$

(D)

$-0.005$

Detailed Solution

Function $f(x) = \frac{2x}{1+x^2}$ , $h=0.5$ . Grid points $x_0=0, x_1=0.5, x_2=1$ . $y_0 = f(0)=0, y_1 = f(0.5) = \frac{1}{1.25} = 0.8$ , $y_2 = f(1)=1$ . Simpson's rule $I \approx \frac{h}{3}(y_0 + 4y_1 + y_2) = \frac{0.5}{3}(0 + 4(0.8) + 1) = \frac{0.5}{3}(4.2) = 0.7$ . Exact integral $I = [\ln(1+x^2)]_0^1 = \ln(2) \approx 0.6931$ . Difference $= 0.7 - 0.6931 = 0.0069 \approx 0.007$ .

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