Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.1 Textbook Questions and Answers, Notes.

## 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 dx 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

The left hand rule for Riemann sum,

S = [f(x_{0}) + f(x_{1})) + f(x_{2}) + f(x_{3}) + f(x_{4})] ∆x

= [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] (0.1)

= 0.6.

Question 2.

Find an approximate value of \(\int_{1}^{1.5}\) x² dx 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) = x²

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_{1} = 1.1

x_{2} = 1.2

x_{3} = 1.3

x_{4} = 1.4

x_{5} = 1.5

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

S = [f(x_{1}) + f(x_{2}) + ….. + f(x_{n})] Δx

∴ S = [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) dx 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

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

The partition of the interval is given by

The partition of the interval is given b.y

x_{0} = 1, x_{1} = 1.1, x_{2}=1.2, x_{3} = 1.3, x_{4} = 1.4, x_{5} = 1.5

The mid – point rule for Riemann sum with equal width Δx is

∴ S = (0.95 + 0.85 + 0.75 + 0.65 + 0.55) × 0.1

= 3.75 × 0.1

= 0.375