PYQs / GATE CE / 2015 / Set 1 / Q12 GATE CE 2015 Set 1 — Question 12 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +1 / -0.33 Medium Trapezoidal Rule Numerical Integration & ODEs Engineering Mathematics
Engineering Mathematics → Numerical Integration & ODEs → Trapezoidal Rule
Last updated 5 September 2026
Question The integral
∫ x 1 x 2 x 2 d x \int_{x_1}^{x_2} x^2 dx ∫ x 1 x 2 x 2 d x with
x 2 > x 1 > 0 x_2 > x_1 > 0 x 2 > x 1 > 0 is evaluated analytically as well as numerically using a single application of the trapezoidal rule. If
I I I is the exact value of the integral obtained analytically and
J J J is the approximate value obtained using the trapezoidal rule, which of the following statements is correct about their relationship?
Solution The function to be integrated is
f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 .
1. Analytical Integration (I I I I = ∫ x 1 x 2 x 2 d x = [ x 3 3 ] x 1 x 2 = x 2 3 − x 1 3 3 I = \int_{x_1}^{x_2} x^2 dx = \left[ \frac{x^3}{3} \right]_{x_1}^{x_2} = \frac{x_2^3 - x_1^3}{3} I = ∫ x 1 x 2 x 2 d x = [ 3 x 3 ] x 1 x 2 = 3 x 2 3 − x 1 3 2. Trapezoidal Rule (J J J For a single application over [ x 1 , x 2 ] [x_1, x_2] [ x 1 , x 2 ] J = x 2 − x 1 2 [ f ( x 1 ) + f ( x 2 ) ] = x 2 − x 1 2 ( x 1 2 + x 2 2 ) J = \frac{x_2 - x_1}{2} [f(x_1) + f(x_2)] = \frac{x_2 - x_1}{2} (x_1^2 + x_2^2) J = 2 x 2 − x 1 [ f ( x 1 ) + f ( x 2 )] = 2 x 2 − x 1 ( x 1 2 + x 2 2 ) Comparing
J J J and
I I I :
J − I = x 2 − x 1 2 ( x 1 2 + x 2 2 ) − x 2 3 − x 1 3 3 J - I = \frac{x_2 - x_1}{2} (x_1^2 + x_2^2) - \frac{x_2^3 - x_1^3}{3} J − I = 2 x 2 − x 1 ( x 1 2 + x 2 2 ) − 3 x 2 3 − x 1 3 Using the identity
x 2 3 − x 1 3 = ( x 2 − x 1 ) ( x 2 2 + x 1 x 2 + x 1 2 ) x_2^3 - x_1^3 = (x_2 - x_1)(x_2^2 + x_1x_2 + x_1^2) x 2 3 − x 1 3 = ( x 2 − x 1 ) ( x 2 2 + x 1 x 2 + x 1 2 ) :
J − I = ( x 2 − x 1 ) [ x 1 2 + x 2 2 2 − x 1 2 + x 1 x 2 + x 2 2 3 ] J - I = (x_2 - x_1) \left[ \frac{x_1^2 + x_2^2}{2} - \frac{x_1^2 + x_1x_2 + x_2^2}{3} \right] J − I = ( x 2 − x 1 ) [ 2 x 1 2 + x 2 2 − 3 x 1 2 + x 1 x 2 + x 2 2 ] J − I = ( x 2 − x 1 ) [ 3 x 1 2 + 3 x 2 2 − 2 x 1 2 − 2 x 1 x 2 − 2 x 2 2 6 ] J - I = (x_2 - x_1) \left[ \frac{3x_1^2 + 3x_2^2 - 2x_1^2 - 2x_1x_2 - 2x_2^2}{6} \right] J − I = ( x 2 − x 1 ) [ 6 3 x 1 2 + 3 x 2 2 − 2 x 1 2 − 2 x 1 x 2 − 2 x 2 2 ] J − I = x 2 − x 1 6 ( x 1 2 − 2 x 1 x 2 + x 2 2 ) = ( x 2 − x 1 ) 3 6 J - I = \frac{x_2 - x_1}{6} (x_1^2 - 2x_1x_2 + x_2^2) = \frac{(x_2 - x_1)^3}{6} J − I = 6 x 2 − x 1 ( x 1 2 − 2 x 1 x 2 + x 2 2 ) = 6 ( x 2 − x 1 ) 3 Since
x 2 > x 1 x_2 > x_1 x 2 > x 1 ,
( x 2 − x 1 ) 3 > 0 (x_2 - x_1)^3 > 0 ( x 2 − x 1 ) 3 > 0 , which means
J − I > 0 ⟹ J > I J - I > 0 \implies J > I J − I > 0 ⟹ J > I .
Alternatively, since
f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 is a concave up function (
f ′ ′ ( x ) = 2 > 0 f''(x) = 2 > 0 f ′′ ( x ) = 2 > 0 ), the trapezoidal rule (which uses a linear chord above the curve) always overestimates the area.
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Solution The function to be integrated is
f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 .
1. Analytical Integration (I I I I = ∫ x 1 x 2 x 2 d x = [ x 3 3 ] x 1 x 2 = x 2 3 − x 1 3 3 I = \int_{x_1}^{x_2} x^2 dx = \left[ \frac{x^3}{3} \right]_{x_1}^{x_2} = \frac{x_2^3 - x_1^3}{3} I = ∫ x 1 x 2 x 2 d x = [ 3 x 3 ] x 1 x 2 = 3 x 2 3 − x 1 3 2. Trapezoidal Rule (J J J For a single application over [ x 1 , x 2 ] [x_1, x_2] [ x 1 , x 2 ] J = x 2 − x 1 2 [ f ( x 1 ) + f ( x 2 ) ] = x 2 − x 1 2 ( x 1 2 + x 2 2 ) J = \frac{x_2 - x_1}{2} [f(x_1) + f(x_2)] = \frac{x_2 - x_1}{2} (x_1^2 + x_2^2) J = 2 x 2 − x 1 [ f ( x 1 ) + f ( x 2 )] = 2 x 2 − x 1 ( x 1 2 + x 2 2 ) Comparing
J J J and
I I I :
J − I = x 2 − x 1 2 ( x 1 2 + x 2 2 ) − x 2 3 − x 1 3 3 J - I = \frac{x_2 - x_1}{2} (x_1^2 + x_2^2) - \frac{x_2^3 - x_1^3}{3} J − I = 2 x 2 − x 1 ( x 1 2 + x 2 2 ) − 3 x 2 3 − x 1 3 Using the identity
x 2 3 − x 1 3 = ( x 2 − x 1 ) ( x 2 2 + x 1 x 2 + x 1 2 ) x_2^3 - x_1^3 = (x_2 - x_1)(x_2^2 + x_1x_2 + x_1^2) x 2 3 − x 1 3 = ( x 2 − x 1 ) ( x 2 2 + x 1 x 2 + x 1 2 ) :
J − I = ( x 2 − x 1 ) [ x 1 2 + x 2 2 2 − x 1 2 + x 1 x 2 + x 2 2 3 ] J - I = (x_2 - x_1) \left[ \frac{x_1^2 + x_2^2}{2} - \frac{x_1^2 + x_1x_2 + x_2^2}{3} \right] J − I = ( x 2 − x 1 ) [ 2 x 1 2 + x 2 2 − 3 x 1 2 + x 1 x 2 + x 2 2 ] J − I = ( x 2 − x 1 ) [ 3 x 1 2 + 3 x 2 2 − 2 x 1 2 − 2 x 1 x 2 − 2 x 2 2 6 ] J - I = (x_2 - x_1) \left[ \frac{3x_1^2 + 3x_2^2 - 2x_1^2 - 2x_1x_2 - 2x_2^2}{6} \right] J − I = ( x 2 − x 1 ) [ 6 3 x 1 2 + 3 x 2 2 − 2 x 1 2 − 2 x 1 x 2 − 2 x 2 2 ] J − I = x 2 − x 1 6 ( x 1 2 − 2 x 1 x 2 + x 2 2 ) = ( x 2 − x 1 ) 3 6 J - I = \frac{x_2 - x_1}{6} (x_1^2 - 2x_1x_2 + x_2^2) = \frac{(x_2 - x_1)^3}{6} J − I = 6 x 2 − x 1 ( x 1 2 − 2 x 1 x 2 + x 2 2 ) = 6 ( x 2 − x 1 ) 3 Since
x 2 > x 1 x_2 > x_1 x 2 > x 1 ,
( x 2 − x 1 ) 3 > 0 (x_2 - x_1)^3 > 0 ( x 2 − x 1 ) 3 > 0 , which means
J − I > 0 ⟹ J > I J - I > 0 \implies J > I J − I > 0 ⟹ J > I .
Alternatively, since
f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 is a concave up function (
f ′ ′ ( x ) = 2 > 0 f''(x) = 2 > 0 f ′′ ( x ) = 2 > 0 ), the trapezoidal rule (which uses a linear chord above the curve) always overestimates the area.
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