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## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.4

Question 1.

If f'(x) = 4x – 5 and f(2) = 1; find f(x)

Solution:

Given f’(x) = 4x – 5 , f(2) = 1

f’ (x) = 4x – 5

\(\frac{\mathrm{d} f(x)}{\mathrm{d} x}\) = 4x – 5

df(x) = (4x – 5) dx

∫ d f(x) = ∫ (4x – 5) dx

f(x) = ∫ 4x . dx – ∫ 5 dx

f(x) = 4∫ x dx – 5∫ dx

f(x) = 4 × \(\frac{x^{2}}{2}\) – 5x + c

f(x) = 2x^{2} – 5x + c ………… (1)

Given f(2) = 1

(1) ⇒ f(2) 2 × 2^{2} – 5 × 2 + e

1 = 2 × 4 – 10 + c

1 = 8 – 10 + c

1 = – 2 + c

c = 1 + 2 = 3

Substituting in equation (1) we get

f(x) = 2x^{2} – 5x + 3

Question 2.

If f'(x) = 9x^{2} – 6x and f(0) = -3; find f(x)

Solution:

f'(x) = \(\int\left(9 x^{2}-6 x\right) d x\)

f(x) = \(\frac{9 x^{3}}{3}-\frac{6 x^{2}}{2}\) + c

f(x) = 3x^{3} – 3x^{2} + c

But f(0) = -3

3(0)^{3} – 3(0)^{2} + c = -3

c = -3

Thus, f(x) = 3x^{3} – 3x^{2} – 3

f(x) = 3(x^{3} – x^{2} – 1)

Question 3.

If f'(x) = 12x – 6 and f(1) = 30, f'(1) = 5 find f(x)

Solution:

f”(x)= 12x – 6 and f(1) = 30, f'(1) = 5

\(\frac{\mathrm{d} f^{\prime}(x)}{\mathrm{d} x}\) = 12x – 6

d f'(x) = (12x – 6) dx

∫ df’ (x) = ∫ (12x – 6) dx

f’ (x) = ∫12x . dx – ∫6 dx

f’ (x) = 12 ∫x dx – 6∫ dx

f’ (x) = 12 × \(\frac{x^{2}}{2}\) 6x + c

f’ (x) = 6x^{2} – 6x + c ………….. (1)

Given f’(1) = 5

(1) ⇒ f’ (1) = 6 × 1^{2} – 6 × 1 + c

5 = 6 – 6 + c

c = 5

Substituting in equation (I) we have

f’ (x) = 6x^{2} – 6x + 5

\(\frac{\mathrm{d} f(x)}{\mathrm{d} x}\) = 6x^{2} – 6x + 5

d f(x) = (6x^{2} – 6x + 5) dx

∫ df(x) = ∫ (6x^{2} – 6x + 5) dx

f(x) = ∫ 6x^{2} . dx – ∫ 6x dx + ∫5 dx

f(x) = 6 ∫ x^{2} . dx – 6 ∫ x dx + 5 ∫ dx

f(x) = 6 × \(\frac{x^{3}}{3}\) – 6 \(\frac{x^{2}}{2}\) + 5x + c_{1}

f(x) = 2x^{3} – 3x^{2} + 5x + c_{1} ……….. (2)

Given f(1) = 30

(2) ⇒ f(1) = 2 × 1^{3} – 3 × 1^{2} + 5 × 1 + c_{1}

30 = 2 – 3 + 5 + c_{1}

30 = 4 + c_{1}

c_{1} = 30 – 4 = 26

Substituting in equation (2) we get

f(x) = 2x^{3} – 3x^{2} + 5x + 26

Question 4.

A ball is thrown vertically upward from the ground with an initial velocity of 39.2 mísec. If the only force considered is that attributed to the acceleration due to gravity, find

(i) how long will it take for the ball to strike the ground?

(ii) the speed with which will it strike the ground? and

(iii) how high the ball will rise?

Solution:

We know that

To find c_{1} by substituting the initial conditions

x = 0 at t = 0 in the above equations,

We get C_{1} = 0

∴ x = \(\frac{a t^{2}}{2}\) + 39.2t ………….. (3)

(i) While the ball strike the ground x = 0

(ii) the speed with which will It strike the ground

To find the speed with which will strike the ground

That is the velocity at t = 8 sec.

s = 39.2 t – 4.9 t^{2}

\(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 39.2 – 4.9 × 2 × t

v = 39.2 – 9.8t

At t = 8, Velocity =. 39.2 – 9.8 × 8

= 39.2 – 78.4

= – 39.2 m/sec.

But velocity cannot be negative.

v = 39.2m/sec.

(iii) how high the ball will rise?

At maximum height velocity = 0

∴ 39.2 – 9.8t = 0

39.2 = 9.8 t

t = \(\frac{39.2}{9.8}\) = 4

t = 4 sec.

∴ Maximum height is obtained by substituting t = 4 in the equation

s = 392 t – 4.9 t^{2}

Maximum height = 39.2 × 4 – 4.9 × 4^{2}

= 156.8 – 4.9 × 16

= 156.8 – 78.4

= 78.4 m.

Question 5.

A wound is healing in such a way that t days since Sunday the area of the wound has been decreasing at a rate of \(\frac{-6}{(t+2)^{2}}\)cm^{2} per day where 0 < r ≤ 8. If on Monday the area of the wound was 1 .4 cm^{2}

(i) What was the area of the wound on Sunday?

(ii) What is the anticipated area of the wound on Thursday if it continues to heal at the

same rate?

Solution:

Let A be the area of the wound at time ‘t’.

A = \(\frac{6}{t+2}\) + c

By the given, condition area of the wound on Monday is 1.4 cm^{2}

⇒ A = 1.4; t = 1

⇒ 1.4 = \(\frac{6}{t+2}\) + c

1.4 = \(\frac{6}{3}\) + c

c = -0.6

∴ Area of wound at any day.

⇒ A = \(\frac{6}{t+2}\) – 0.6

(i) The area of a wound on Sunday

t = 0 ⇒ A = \(\frac{6}{2}\) – 0.6 = 3 – 0.6 = 2.4 cm^{2}

(ii) The area of the wound on Thrusday

t = 4 ⇒ A = \(\frac{6}{6}\) – 0.6 = 1 – 0.6 = 0.4 cm^{2}

### Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.4 Additional Problems

Question 1.

If f'(x) = 2x – 7 and f(1) = 0 find f(x)

Solution:

Given f’ (x) = 2x – 7

⇒ f (x) = \(\int(2 x-7) d x\)

= x^{2} – 7x + c

Given f(1) = 0 ⇒ 1 – 7 + c = 0

⇒ c = 6

So f(x) = x^{2} – 7x + 6

Question 2.

Given f”(x) = 6x + 6, f'(0) = -5 and f(1) = 6 find f(x)

Solution:

f” (x) = 6x + 6

⇒ f ‘ (x) = (6x + 6) dx

= \(\frac{6 x^{2}}{2}\) + 6x + c

= 3x^{2} + 6x + c

Given f ‘(0) = -5 ⇒ 0 + c = -5

⇒ c = -5

∴ f ‘(x) = 3x^{2} + 6x – 5

So f(x) = \(\int\left(3 x^{2}+6 x-5\right) d x\)

= \(\frac{3 x^{3}}{3}+\frac{6 x^{2}}{2}\) – 5x + c

(i.e.,) f(x) = x^{3} + 3x^{2} – 5x + c

Given f(1) = 6

⇒ 1 + 3 + 5 + c = 6

⇒ c – 1 = 6 ⇒ c = 7

So f(x) = x^{3} + 3x^{2} – 5x + 7