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## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.2

Question 1.

Evaluate the following integrals as the limits of sums:

Solution:

We use the formula

(ii) \(\int_{1}^{2}\left(4 x^{2}-1\right) d x\)

Solution:

We use the formula

### Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.2 Additional Problems

Question 1.

Evaluate as the limit of sums:

Solution:

Let f(x) = 2x^{2} +5; a = 1; b = 3 and nh = 3 – 1 = 2

Here, f(a) = f(1) = 2(1)^{2} + 5

f(a + h) = f(1 + h) = 2(1 + h)^{2} + 5

f(a + 2h) = f(1 + 2h) = 2 (1 + 2h)^{2} + 5

f[a+(n – 1)h = f[1 + (n – 1)h] = 2[1 + (n – 1)h]^{2} + 5

Question 2.

Evaluates as the limit of sums: \(\int_{1}^{2}\left(x^{2}-1\right) d x\)

Solution:

Let f(x) = x^{2} – 1 for 1 ≤ x ≤ 2

We divide the interval [1, 2] into n equal sub-intervals each of length h.

We have a = 1, b = 2

Here, f(a) = f(1) = (1)^{2} – 1 = 0

f(a + h) = f(1 + h) = (1 + h)^{2} – 1 = 1 + h^{2} + 2h – 1 = h^{2} + 2h

f(a + 2h) = f(1 + 2h) = (1 + 2h)^{2} – 1 = 1 + 4h^{2} + 4h – 1 = 4h^{2} + 4h

f[a + (n -1)h] = f[1 + (n – 1)h] = [1 + (n – 1)h]^{2} – 1

= 1 + (n – 1)^{2} h^{2} + 2(n – 1 )h – 1 = (n – 1)^{2} h^{2} + 2 (n – 1 )h

Question 3.

Evalute:

Solution:

Let f(x) = x^{2} + x + 2, for 0 ≤ x ≤ 2 Here, a = 0,b = 2

We divide the closed interval [0, 2] into n subintervals each of length h.

By definition,