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## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.1

Question 1.

Find an approximate value of \(\int_{1}^{1.5} x d x\) by applying the left-end rule with the partition {1.1, 1.2, 1.3, 1.4, 1.5}

Solution:

Here a = 1, b = 1.5, n = 5, f(x) = x

So, the width of each subinterval is

h = Δx = \(\frac{b-a}{n}\) = \(\frac{1.5-1}{5}\) = 0.1

The partition of the interval is given by

x_{0} = 1

x_{1} = x_{0} + h = 1 + 0.1 = 1.1

x_{2} = x_{1} + h = 1.1 + 0.1 = 1.2

x_{3} = x_{2} + h = 1.2 + 0.1 = 1.3

x_{4} = x_{3} + h = 1.3 + 0.1 = 1.4

x_{5} = x_{4} + h = 1.4 + 0.1 = 1.5

The Left – end rule for Riemann sum with equal width Δx is

S = [f(x_{0}) + f(x_{1}) + ….. + f(x_{n-1})] Δx

∴ S = [f(1) + f(1.1) + f(1.2) + f(1.3) + f(1.4)](0.1)

= (1 + 1.1 + 1.2 + 1.3 + 1.4) × 0.1

= 6.0 × 0.1

S = 0.6

Question 2.

Find an approximate value of \(\int_{1}^{1.5} x^{2} d x\) by applying the right-end rule with the partition {1.1, 1.2, 1.3, 1.4, 1.5}.

Solution:

Here a = 1;

b = 1.5;

n = 5;

f(x) = x^{2}

So, the width of each subinterval is

The Right hand rule for Riemann sum,

S = [f(x_{1}) +f(x_{2}) +f(x_{3}) + f(x_{4}) + f(x_{5})] ∆x

= [f(1.1) + f(1.2) + f(1.3) + f(1.4) + f(1.5)] (0.1)

= [1.21 + 1.44 + 1.69 + 1.96 + 2.25] (0.1)

= [8.55] (0.1)

= 0.855.

Question 3.

Find an approximate value of \(\int_{1}^{1.5}(2-x) d x\) by applying the mid-point rule with the partition {1.1, 1.2, 1.3, 1.4, 1.5}.

Solution:

Here a = 1;

b = 1.5;

n = 5;

f(x) = 2 – x

So, the width of each subinterval is

The mid-point rule for Riemann sum,

= [f(1.05) + f(1 .15) + f(1.25) +f(1.35) + f(1 .45)] (0.1)

= [0.95 + 0.85 + 0.75 + 0.65 + 0.55] (0.1)

= [3.75] (0.1)

= 0.375.