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Integrals

7.7. Integrals of Some More Types.

Q.1. Evaluate : ∫√(4 – x2) dx.

Solution :

We have, ∫√(4 – x²) dx = ∫1.√(4 – x²) dx
= x.(1/2)(4 – x2)–1/2 (– 2x).dx [Integrating by parts]
= x.√(4 – x²) – ∫(– x²)/√(4 – x²) dx
= x.√(4 – x²) – ∫{(4 – x²) – 4 }/√(4 – x²) dx
= x.√(4 – x²) – ∫√(4 – x²) dx + 4∫dx/√(4 – x²)
Or, 2∫√(4 – x²) dx = x.√(4 – x²) + 4∫dx/√(22 – x²)
= x.√(4 –x²) + 4 sin –1(x/2)
Therefore, ∫√(4 – x2) dx = (1/2). x.√(4 – x2) + 2 sin –1(x/2) + C. [Ans.]

Q.2. Evaluate : ∫[(x² + 1)/(x4 + 1)] dx.

Solution :

We have, I = ∫[(x² + 1)/(x⁴ + 1)] dx
= ∫[(1 + 1/x²)/(x² + 1/x²)] dx
Put x – 1/x = t, then (1 + 1/x²)dx = dt
Also x² + 1/x² = (x – 1/x)² + 2 = t² + 2.
Therefore, I = ∫dt/(t² + 2) = ∫dt/{t² + (√2)²}
= (1/√2)tan^-1 (t/√2) + c
= (1/√2)tan ^-1 [(x – 1/x)/√2] + c
= tan –1 [(x² – 1)/(√2)x] + c. [Ans.]

7.8. Definite Integral as The Limit of a Sum.

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

Solution :

Let I = 0∫2 (x² + x + 2) dx .
Here, we have, f(x) = x² + x + 2; a = 0, b = 2
Therefore, n h = b – a = 2.
By definition,
a∫b f(x) dx = lim h→0 h[ f(a) + f(a + h) + ------- + f{a + (n – 1)h }]
where nh = b – a .
Therefore, I = lim h→0 h[(0² + 0 + 2) + (h² + h + 2) + ---- + (n – 1)2h² + (n – 1)h + 2]
where nh = 2.
= lim h→0 h[{1² + 2² + ------ + (n – 1)2}h² + {1 + 2 + ------ (n – 1)}h + 2n]
= lim h→0 h[1/6(n – 1)n(2n – 1)h2 + {(n – 1)n/2}h + 2n}
= lim h→0 [1/6(nh – h)(nh)(2nh – h) + 1/2 (2 – h).2 + 2×2]
= 1/6 (2)(2)(4) + 1/2 (2)(2) + 4
= 26/3. [Ans.]

Paper By Mr. M.P.Keshari
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