PYQs / GATE CE / 2021 / Set 2 / Q56 GATE CE 2021 Set 2 — Question 56 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 Trapezoidal Rule Numerical Integration & ODEs Engineering Mathematics
Engineering Mathematics → Numerical Integration & ODEs → Trapezoidal Rule
Last updated 5 September 2026
Question Numerically integrate,
f ( x ) = 10 x − 20 x 2 f(x) = 10x - 20x^2 f ( x ) = 10 x − 20 x 2 from lower limit
a = 0 a = 0 a = 0 to upper limit
b = 0.5 b = 0.5 b = 0.5 . Use Trapezoidal rule with five equal subdivisions. The value (in units, round off to two decimal places) obtained is
________ .
Correct answer 0.38 to 0.42
Solution Given:
f ( x ) = 10 x − 20 x 2 f(x) = 10x - 20x^2 f ( x ) = 10 x − 20 x 2 Lower limit
a = 0 a = 0 a = 0 , Upper limit
b = 0.5 b = 0.5 b = 0.5 Number of subdivisions
n = 5 n = 5 n = 5 Step size
h = b − a n = 0.5 − 0 5 = 0.1 h = \frac{b - a}{n} = \frac{0.5 - 0}{5} = 0.1 h = n b − a = 5 0.5 − 0 = 0.1 The values of
x x x and
y = f ( x ) y = f(x) y = f ( x ) are:
x 0 = 0 , y 0 = f ( 0 ) = 10 ( 0 ) − 20 ( 0 ) 2 = 0 x_0 = 0, y_0 = f(0) = 10(0) - 20(0)^2 = 0 x 0 = 0 , y 0 = f ( 0 ) = 10 ( 0 ) − 20 ( 0 ) 2 = 0 x 1 = 0.1 , y 1 = f ( 0.1 ) = 10 ( 0.1 ) − 20 ( 0.1 ) 2 = 1 − 0.2 = 0.8 x_1 = 0.1, y_1 = f(0.1) = 10(0.1) - 20(0.1)^2 = 1 - 0.2 = 0.8 x 1 = 0.1 , y 1 = f ( 0.1 ) = 10 ( 0.1 ) − 20 ( 0.1 ) 2 = 1 − 0.2 = 0.8 x 2 = 0.2 , y 2 = f ( 0.2 ) = 10 ( 0.2 ) − 20 ( 0.2 ) 2 = 2 − 0.8 = 1.2 x_2 = 0.2, y_2 = f(0.2) = 10(0.2) - 20(0.2)^2 = 2 - 0.8 = 1.2 x 2 = 0.2 , y 2 = f ( 0.2 ) = 10 ( 0.2 ) − 20 ( 0.2 ) 2 = 2 − 0.8 = 1.2 x 3 = 0.3 , y 3 = f ( 0.3 ) = 10 ( 0.3 ) − 20 ( 0.3 ) 2 = 3 − 1.8 = 1.2 x_3 = 0.3, y_3 = f(0.3) = 10(0.3) - 20(0.3)^2 = 3 - 1.8 = 1.2 x 3 = 0.3 , y 3 = f ( 0.3 ) = 10 ( 0.3 ) − 20 ( 0.3 ) 2 = 3 − 1.8 = 1.2 x 4 = 0.4 , y 4 = f ( 0.4 ) = 10 ( 0.4 ) − 20 ( 0.4 ) 2 = 4 − 3.2 = 0.8 x_4 = 0.4, y_4 = f(0.4) = 10(0.4) - 20(0.4)^2 = 4 - 3.2 = 0.8 x 4 = 0.4 , y 4 = f ( 0.4 ) = 10 ( 0.4 ) − 20 ( 0.4 ) 2 = 4 − 3.2 = 0.8 x 5 = 0.5 , y 5 = f ( 0.5 ) = 10 ( 0.5 ) − 20 ( 0.5 ) 2 = 5 − 5 = 0 x_5 = 0.5, y_5 = f(0.5) = 10(0.5) - 20(0.5)^2 = 5 - 5 = 0 x 5 = 0.5 , y 5 = f ( 0.5 ) = 10 ( 0.5 ) − 20 ( 0.5 ) 2 = 5 − 5 = 0 Using the Trapezoidal rule:
I = h 2 [ y 0 + y 5 + 2 ( y 1 + y 2 + y 3 + y 4 ) ] I = \frac{h}{2} [y_0 + y_5 + 2(y_1 + y_2 + y_3 + y_4)] I = 2 h [ y 0 + y 5 + 2 ( y 1 + y 2 + y 3 + y 4 )] I = 0.1 2 [ 0 + 0 + 2 ( 0.8 + 1.2 + 1.2 + 0.8 ) ] I = \frac{0.1}{2} [0 + 0 + 2(0.8 + 1.2 + 1.2 + 0.8)] I = 2 0.1 [ 0 + 0 + 2 ( 0.8 + 1.2 + 1.2 + 0.8 )] I = 0.05 [ 2 ( 4.0 ) ] = 0.05 × 8 = 0.40 I = 0.05 [2(4.0)] = 0.05 \times 8 = 0.40 I = 0.05 [ 2 ( 4.0 )] = 0.05 × 8 = 0.40 The value obtained is 0.40.
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Correct answer 0.38 to 0.42
Solution Given:
f ( x ) = 10 x − 20 x 2 f(x) = 10x - 20x^2 f ( x ) = 10 x − 20 x 2 Lower limit
a = 0 a = 0 a = 0 , Upper limit
b = 0.5 b = 0.5 b = 0.5 Number of subdivisions
n = 5 n = 5 n = 5 Step size
h = b − a n = 0.5 − 0 5 = 0.1 h = \frac{b - a}{n} = \frac{0.5 - 0}{5} = 0.1 h = n b − a = 5 0.5 − 0 = 0.1 The values of
x x x and
y = f ( x ) y = f(x) y = f ( x ) are:
x 0 = 0 , y 0 = f ( 0 ) = 10 ( 0 ) − 20 ( 0 ) 2 = 0 x_0 = 0, y_0 = f(0) = 10(0) - 20(0)^2 = 0 x 0 = 0 , y 0 = f ( 0 ) = 10 ( 0 ) − 20 ( 0 ) 2 = 0 x 1 = 0.1 , y 1 = f ( 0.1 ) = 10 ( 0.1 ) − 20 ( 0.1 ) 2 = 1 − 0.2 = 0.8 x_1 = 0.1, y_1 = f(0.1) = 10(0.1) - 20(0.1)^2 = 1 - 0.2 = 0.8 x 1 = 0.1 , y 1 = f ( 0.1 ) = 10 ( 0.1 ) − 20 ( 0.1 ) 2 = 1 − 0.2 = 0.8 x 2 = 0.2 , y 2 = f ( 0.2 ) = 10 ( 0.2 ) − 20 ( 0.2 ) 2 = 2 − 0.8 = 1.2 x_2 = 0.2, y_2 = f(0.2) = 10(0.2) - 20(0.2)^2 = 2 - 0.8 = 1.2 x 2 = 0.2 , y 2 = f ( 0.2 ) = 10 ( 0.2 ) − 20 ( 0.2 ) 2 = 2 − 0.8 = 1.2 x 3 = 0.3 , y 3 = f ( 0.3 ) = 10 ( 0.3 ) − 20 ( 0.3 ) 2 = 3 − 1.8 = 1.2 x_3 = 0.3, y_3 = f(0.3) = 10(0.3) - 20(0.3)^2 = 3 - 1.8 = 1.2 x 3 = 0.3 , y 3 = f ( 0.3 ) = 10 ( 0.3 ) − 20 ( 0.3 ) 2 = 3 − 1.8 = 1.2 x 4 = 0.4 , y 4 = f ( 0.4 ) = 10 ( 0.4 ) − 20 ( 0.4 ) 2 = 4 − 3.2 = 0.8 x_4 = 0.4, y_4 = f(0.4) = 10(0.4) - 20(0.4)^2 = 4 - 3.2 = 0.8 x 4 = 0.4 , y 4 = f ( 0.4 ) = 10 ( 0.4 ) − 20 ( 0.4 ) 2 = 4 − 3.2 = 0.8 x 5 = 0.5 , y 5 = f ( 0.5 ) = 10 ( 0.5 ) − 20 ( 0.5 ) 2 = 5 − 5 = 0 x_5 = 0.5, y_5 = f(0.5) = 10(0.5) - 20(0.5)^2 = 5 - 5 = 0 x 5 = 0.5 , y 5 = f ( 0.5 ) = 10 ( 0.5 ) − 20 ( 0.5 ) 2 = 5 − 5 = 0 Using the Trapezoidal rule:
I = h 2 [ y 0 + y 5 + 2 ( y 1 + y 2 + y 3 + y 4 ) ] I = \frac{h}{2} [y_0 + y_5 + 2(y_1 + y_2 + y_3 + y_4)] I = 2 h [ y 0 + y 5 + 2 ( y 1 + y 2 + y 3 + y 4 )] I = 0.1 2 [ 0 + 0 + 2 ( 0.8 + 1.2 + 1.2 + 0.8 ) ] I = \frac{0.1}{2} [0 + 0 + 2(0.8 + 1.2 + 1.2 + 0.8)] I = 2 0.1 [ 0 + 0 + 2 ( 0.8 + 1.2 + 1.2 + 0.8 )] I = 0.05 [ 2 ( 4.0 ) ] = 0.05 × 8 = 0.40 I = 0.05 [2(4.0)] = 0.05 \times 8 = 0.40 I = 0.05 [ 2 ( 4.0 )] = 0.05 × 8 = 0.40 The value obtained is 0.40.
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