CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Integrals

Q.2. Evaluate : 0∫2(x² + x + 1) dx as limit of a sum.

Solution :

We have, f(x) = x² + x + 1, a = 0, b = 2 => n h = b – a = 2 – 0 = 2.
By the formula,
a∫bf(x) dx = limh→0 h[f(a) + f(a + h) + ----- + f{a + (n – 1)h}]
where nh = b – a
Therefore, 0∫2(x² + x + 1) dx = limh→0 h[(0² + 0 + 1) + (h² + h + 1) + {(2h)² + 2h + 1}
+ ----- + {(n – 1)2h² + (n – 1)h + 1] where nh = 2.
= limh→0 h[{12 + 22 + ---- + (n – 1)2}h2 + {1 + 2 + ----- + (n – 1)}h + n]
where nh = 2.
= limh→0 h[1/6(n – 1)n(2n – 1)h² + 1/2n(n – 1)h + n], where nh = 2.
= limh→0 [1/6(nh – h)nh(2nh – h) + 1/2(nh – 1)nh + nh], where, nh = 2.
= 1/6(2 – 0)×2(2×2 – 0) + 1/2(2 – 0)×2 + 2
= 8/3 + 2 + 2
= 20/3. [Ans.]

Q.3. Evaluate : 1∫3(x² + 5x)dx as the limit of sums.

Solution :

o yourself. [Ans. = 86/3.]

Q.4. Evaluate : 1∫3(x² + x)dx as limit of sums.

Solution :

Do yourself. [Ans. = 38/3.]

7.9. Evaluation of Definite Integrals.

Q.1 Evaluate : ₀∫¹ dx/(1 + x²).

Solution :

We have, ₀∫¹ dx/(1 + x²) = I (say)
Put x = tan θ, then dx = sec² θdθ and 1 + x² = 1 + tan2 θ = sec2 θ.
When x = 0 , tan θ = 0 => θ = 0 and when x = 1, θ = π/4.
Therefore, I = ₀∫^π/4sec2 θdθ/sec2θ
= ₀∫^π/4dθ
= [θ]0π/4 = π/4. [Ans.]

Alternative

Let I = 0∫1dx/(1 +x²)
= [tan –1 x]01
= tan –1 (1) – tan –1 (0)
= π/4 – 0 = π/4. [Ans.]

Paper By Mr. M.P.Keshari
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