PYQs / GATE CE / 2015 / Set 2 / Q39 GATE CE 2015 Set 2 — Question 39 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. NAT +2 / -0 Medium Simpson's Rule Numerical Integration & ODEs Engineering Mathematics
Engineering Mathematics → Numerical Integration & ODEs → Simpson's Rule
Last updated 5 September 2026
Question For step-size,
Δ x = 0.4 \Delta x = 0.4 Δ x = 0.4 , the value of following integral using Simpson's 1/3 rule is
__________ ∫ 0 0.8 ( 0.2 + 25 x − 200 x 2 + 675 x 3 − 900 x 4 + 400 x 5 ) d x \int_{0}^{0.8} (0.2 + 25x - 200x^2 + 675x^3 - 900x^4 + 400x^5) dx ∫ 0 0.8 ( 0.2 + 25 x − 200 x 2 + 675 x 3 − 900 x 4 + 400 x 5 ) d x Correct answer 1.36 to 1.37
Solution Given
h = Δ x = 0.4 h = \Delta x = 0.4 h = Δ x = 0.4 and the interval
[ 0 , 0.8 ] [0, 0.8] [ 0 , 0.8 ] . The number of intervals
n = 0.8 − 0 0.4 = 2 n = \frac{0.8 - 0}{0.4} = 2 n = 0.4 0.8 − 0 = 2 .
The points are
x 0 = 0 , x 1 = 0.4 , x 2 = 0.8 x_0 = 0, x_1 = 0.4, x_2 = 0.8 x 0 = 0 , x 1 = 0.4 , x 2 = 0.8 .
Let
f ( x ) = 0.2 + 25 x − 200 x 2 + 675 x 3 − 900 x 4 + 400 x 5 f(x) = 0.2 + 25x - 200x^2 + 675x^3 - 900x^4 + 400x^5 f ( x ) = 0.2 + 25 x − 200 x 2 + 675 x 3 − 900 x 4 + 400 x 5 .
y 0 = f ( 0 ) = 0.2 y_0 = f(0) = 0.2 y 0 = f ( 0 ) = 0.2 y 1 = f ( 0.4 ) = 0.2 + 25 ( 0.4 ) − 200 ( 0.4 ) 2 + 675 ( 0.4 ) 3 − 900 ( 0.4 ) 4 + 400 ( 0.4 ) 5 = 2.456 y_1 = f(0.4) = 0.2 + 25(0.4) - 200(0.4)^2 + 675(0.4)^3 - 900(0.4)^4 + 400(0.4)^5 = 2.456 y 1 = f ( 0.4 ) = 0.2 + 25 ( 0.4 ) − 200 ( 0.4 ) 2 + 675 ( 0.4 ) 3 − 900 ( 0.4 ) 4 + 400 ( 0.4 ) 5 = 2.456 y 2 = f ( 0.8 ) = 0.2 + 25 ( 0.8 ) − 200 ( 0.8 ) 2 + 675 ( 0.8 ) 3 − 900 ( 0.8 ) 4 + 400 ( 0.8 ) 5 = 0.232 y_2 = f(0.8) = 0.2 + 25(0.8) - 200(0.8)^2 + 675(0.8)^3 - 900(0.8)^4 + 400(0.8)^5 = 0.232 y 2 = f ( 0.8 ) = 0.2 + 25 ( 0.8 ) − 200 ( 0.8 ) 2 + 675 ( 0.8 ) 3 − 900 ( 0.8 ) 4 + 400 ( 0.8 ) 5 = 0.232 Using Simpson's 1/3 rule:
I = h 3 [ y 0 + 4 y 1 + y 2 ] I = \frac{h}{3} [y_0 + 4y_1 + y_2] I = 3 h [ y 0 + 4 y 1 + y 2 ] I = 0.4 3 [ 0.2 + 4 ( 2.456 ) + 0.232 ] I = \frac{0.4}{3} [0.2 + 4(2.456) + 0.232] I = 3 0.4 [ 0.2 + 4 ( 2.456 ) + 0.232 ] I = 0.4 3 [ 0.2 + 9.824 + 0.232 ] = 0.4 3 [ 10.256 ] ≈ 1.36747 I = \frac{0.4}{3} [0.2 + 9.824 + 0.232] = \frac{0.4}{3} [10.256] \approx 1.36747 I = 3 0.4 [ 0.2 + 9.824 + 0.232 ] = 3 0.4 [ 10.256 ] ≈ 1.36747 The value lies in the range 1.36 to 1.37.
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Correct answer 1.36 to 1.37
Solution Given
h = Δ x = 0.4 h = \Delta x = 0.4 h = Δ x = 0.4 and the interval
[ 0 , 0.8 ] [0, 0.8] [ 0 , 0.8 ] . The number of intervals
n = 0.8 − 0 0.4 = 2 n = \frac{0.8 - 0}{0.4} = 2 n = 0.4 0.8 − 0 = 2 .
The points are
x 0 = 0 , x 1 = 0.4 , x 2 = 0.8 x_0 = 0, x_1 = 0.4, x_2 = 0.8 x 0 = 0 , x 1 = 0.4 , x 2 = 0.8 .
Let
f ( x ) = 0.2 + 25 x − 200 x 2 + 675 x 3 − 900 x 4 + 400 x 5 f(x) = 0.2 + 25x - 200x^2 + 675x^3 - 900x^4 + 400x^5 f ( x ) = 0.2 + 25 x − 200 x 2 + 675 x 3 − 900 x 4 + 400 x 5 .
y 0 = f ( 0 ) = 0.2 y_0 = f(0) = 0.2 y 0 = f ( 0 ) = 0.2 y 1 = f ( 0.4 ) = 0.2 + 25 ( 0.4 ) − 200 ( 0.4 ) 2 + 675 ( 0.4 ) 3 − 900 ( 0.4 ) 4 + 400 ( 0.4 ) 5 = 2.456 y_1 = f(0.4) = 0.2 + 25(0.4) - 200(0.4)^2 + 675(0.4)^3 - 900(0.4)^4 + 400(0.4)^5 = 2.456 y 1 = f ( 0.4 ) = 0.2 + 25 ( 0.4 ) − 200 ( 0.4 ) 2 + 675 ( 0.4 ) 3 − 900 ( 0.4 ) 4 + 400 ( 0.4 ) 5 = 2.456 y 2 = f ( 0.8 ) = 0.2 + 25 ( 0.8 ) − 200 ( 0.8 ) 2 + 675 ( 0.8 ) 3 − 900 ( 0.8 ) 4 + 400 ( 0.8 ) 5 = 0.232 y_2 = f(0.8) = 0.2 + 25(0.8) - 200(0.8)^2 + 675(0.8)^3 - 900(0.8)^4 + 400(0.8)^5 = 0.232 y 2 = f ( 0.8 ) = 0.2 + 25 ( 0.8 ) − 200 ( 0.8 ) 2 + 675 ( 0.8 ) 3 − 900 ( 0.8 ) 4 + 400 ( 0.8 ) 5 = 0.232 Using Simpson's 1/3 rule:
I = h 3 [ y 0 + 4 y 1 + y 2 ] I = \frac{h}{3} [y_0 + 4y_1 + y_2] I = 3 h [ y 0 + 4 y 1 + y 2 ] I = 0.4 3 [ 0.2 + 4 ( 2.456 ) + 0.232 ] I = \frac{0.4}{3} [0.2 + 4(2.456) + 0.232] I = 3 0.4 [ 0.2 + 4 ( 2.456 ) + 0.232 ] I = 0.4 3 [ 0.2 + 9.824 + 0.232 ] = 0.4 3 [ 10.256 ] ≈ 1.36747 I = \frac{0.4}{3} [0.2 + 9.824 + 0.232] = \frac{0.4}{3} [10.256] \approx 1.36747 I = 3 0.4 [ 0.2 + 9.824 + 0.232 ] = 3 0.4 [ 10.256 ] ≈ 1.36747 The value lies in the range 1.36 to 1.37.
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