CBSE Class 11 Physics Calculus for Physics Notes Set 02

Practice Questions

Functions

Question 1. If \( f(x) = \frac{x - 1}{x + 1} \), then find f{f(x)}.
Answer:
\( f(f(x)) = \frac{\frac{x - 1}{x + 1} - 1}{\frac{x - 1}{x + 1} + 1} = \frac{(x - 1) - (x + 1)}{(x - 1) + (x + 1)} = \frac{-2}{2x} \)
\( \therefore f\{f(x)\} = -\frac{1}{x} \)

Question 2. If f(x) = x + 2 for x < 2 and f(x) = 2x − 1 for x ≥ 2, evaluate f(2), f(1) and f(3).
Answer:
f(2): since 2 ≥ 2, f(2) = 2(2) − 1 = 3
f(1): since 1 < 2, f(1) = 1 + 2 = 3
f(3): since 3 ≥ 2, f(3) = 2(3) − 1 = 5

Differentiation of Elementary Functions

Find the derivative of the given function with respect to the corresponding independent variable.

Question 3. y = x² + x + 8
Answer:
\( \frac{dy}{dx} = 2x + 1 + 0 = 2x + 1 \)

Question 4. s = 5t³ − 3t⁵
Answer:
\( \frac{ds}{dt} = 5(3t^2) - 3(5t^4) = 15t^2 - 15t^4 \)

Question 5. y = 5 sin x
Answer:
\( \frac{dy}{dx} = 5\frac{d}{dx}(\sin x) = 5\cos x \)

Question 6. y = x² + sin x
Answer:
\( \frac{dy}{dx} = 2x + \cos x \)

Question 7. y = tan x + cot x
Answer:
\( \frac{dy}{dx} = \sec^2 x - \text{cosec}^2 x \)

Find the first derivative and second derivative of the given function with respect to the corresponding independent variable.

Question 8. y = 6x² − 10x − 5x⁻²
Answer:
\( \frac{dy}{dx} = 12x - 10 - 5(-2)x^{-3} = 12x - 10 + 10x^{-3} \)
\( \frac{d^2y}{dx^2} = 12 + 10(-3)x^{-4} = 12 - 30x^{-4} \)

Question 9. \( r = \frac{12}{\theta} - \frac{4}{\theta^3} + \frac{1}{\theta^4} \)
Answer:
r = 12θ⁻¹ − 4θ⁻³ + θ⁻⁴
\( \frac{dr}{d\theta} = -12\theta^{-2} + 12\theta^{-4} - 4\theta^{-5} \)
\( \frac{d^2r}{d\theta^2} = 24\theta^{-3} - 48\theta^{-5} + 20\theta^{-6} \)

Question 10. ω = 3z⁷ − 7z³ + 21z²
Answer:
\( \frac{d\omega}{dz} = 21z^6 - 21z^2 + 42z \)
\( \frac{d^2\omega}{dz^2} = 126z^5 - 42z + 42 \)

Question 11. y = sin x + cos x
Answer:
\( \frac{dy}{dx} = \cos x - \sin x \)
\( \frac{d^2y}{dx^2} = -\sin x - \cos x \)

Question 12. y = ln x + e^x
Answer:
\( \frac{dy}{dx} = \frac{1}{x} + e^x \)
\( \frac{d^2y}{dx^2} = -\frac{1}{x^2} + e^x \)

Question 13. y = ln x² + sin x
Answer:
y = 2 ln x + sin x (for x > 0; the result holds for all x ≠ 0)
\( \frac{dy}{dx} = \frac{2}{x} + \cos x \)
\( \frac{d^2y}{dx^2} = -\frac{2}{x^2} - \sin x \)

Question 14. \( y = \sqrt[7]{x} + \tan x \)
Answer:
y = x^1/7 + tan x
\( \frac{dy}{dx} = \frac{1}{7}x^{-6/7} + \sec^2 x \)
\( \frac{d^2y}{dx^2} = -\frac{6}{49}x^{-13/7} + 2\sec x\cdot\sec x\tan x = -\frac{6}{49}x^{-13/7} + 2\tan x\sec^2 x \)

Differentiation by Product Rule

Find the derivative of the given function with respect to the independent variable.

Question 15. y = x sin x
Answer:
\( \frac{dy}{dx} = x\cos x + \sin x\cdot 1 = \sin x + x\cos x \)

Question 16. y = e^x ln x
Answer:
\( \frac{dy}{dx} = e^x\cdot\frac{1}{x} + \ln x\cdot e^x = e^x\ln x + \frac{e^x}{x} \)

Question 17. y = (x − 1)(x² + x + 1)
Answer:
\( \frac{dy}{dx} = (x - 1)(2x + 1) + (x^2 + x + 1)(1) = 2x^2 - x - 1 + x^2 + x + 1 = 3x^2 \)
(Check: y = x³ − 1, so dy/dx = 3x².)

Question 18. \( y = (x^2 + 1)\left(x + 5 + \frac{1}{x}\right) \)
Answer:
\( y' = (x^2 + 1)\left(1 - \frac{1}{x^2}\right) + \left(x + 5 + \frac{1}{x}\right)(2x) \)
\( = x^2 - 1 + 1 - \frac{1}{x^2} + 2x^2 + 10x + 2 = 3x^2 + 10x + 2 - \frac{1}{x^2} \)

Question 19. y = sin x cos x
Answer:
\( \frac{dy}{dx} = \sin x(-\sin x) + \cos x(\cos x) = \cos^2 x - \sin^2 x \) (= cos 2x)

Question 20. y = x² cot x
Answer:
\( \frac{dy}{dx} = x^2(-\text{cosec}^2 x) + \cot x(2x) = -x^2\,\text{cosec}^2 x + 2x\cot x \)

Question 21. y = e^x tan x
Answer:
\( \frac{dy}{dx} = e^x\sec^2 x + \tan x\cdot e^x = e^x(\tan x + \sec^2 x) \)

Question 22. y = x² sin⁴ x + x cos⁻² x
Answer:
\( \frac{d}{dx}(x^2\sin^4 x) = 2x\sin^4 x + x^2\cdot 4\sin^3 x\cos x \)
\( \frac{d}{dx}(x\cos^{-2} x) = \cos^{-2} x + x(-2)\cos^{-3} x(-\sin x) = \cos^{-2} x + 2x\cos^{-3} x\sin x \)
\( \therefore \frac{dy}{dx} = 2x\sin^4 x + 4x^2\sin^3 x\cos x + \cos^{-2} x + 2x\cos^{-3} x\sin x \)

Question 23. \( y = \left(x + \frac{1}{x}\right)\left(x - \frac{1}{x} + 1\right) \)
Answer:
\( \frac{dy}{dx} = \left(1 - \frac{1}{x^2}\right)\left(x - \frac{1}{x} + 1\right) + \left(x + \frac{1}{x}\right)\left(1 + \frac{1}{x^2}\right) \)
\( = \left(x - \frac{1}{x} + 1 - \frac{1}{x} + \frac{1}{x^3} - \frac{1}{x^2}\right) + \left(x + \frac{1}{x} + \frac{1}{x} + \frac{1}{x^3}\right) \)
\( = 1 + 2x + \frac{2}{x^3} - \frac{1}{x^2} \)

Question 24. y = x² sin x + 2x cos x − 2 sin x
Answer:
\( \frac{dy}{dx} = (x^2\cos x + 2x\sin x) + (-2x\sin x + 2\cos x) - 2\cos x \)
\( = x^2\cos x \)

Question 25. y = x² cos x − 2x sin x − 2 cos x
Answer:
\( \frac{dy}{dx} = (-x^2\sin x + 2x\cos x) - (2x\cos x + 2\sin x) + 2\sin x \)
\( = -x^2\sin x \)

Question 26. r = (1 + sec θ) sin θ
Answer:
\( \frac{dr}{d\theta} = (1 + \sec\theta)\cos\theta + \sin\theta\cdot\sec\theta\tan\theta = \cos\theta + 1 + \tan^2\theta \)
\( = \cos\theta + \sec^2\theta \)

Differentiation by Quotient Rule

Find the derivative of the given function with respect to the independent variable.

Question 27. \( y = \frac{\sin x}{\cos x} \)
Answer:
\( \frac{dy}{dx} = \frac{\cos x\cos x - \sin x(-\sin x)}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x \)

Question 28. \( y = \frac{2x + 5}{3x - 2} \)
Answer:
\( y' = \frac{(3x - 2)(2) - (2x + 5)(3)}{(3x - 2)^2} = \frac{6x - 4 - 6x - 15}{(3x - 2)^2} = \frac{-19}{(3x - 2)^2} \)

Question 29. \( y = \frac{\ln x}{x} \)
Answer:
\( \frac{dy}{dx} = \frac{x\cdot\frac{1}{x} - \ln x\cdot 1}{x^2} = \frac{1}{x^2} - \frac{\ln x}{x^2} \)

Question 30. \( f(t) = \frac{t^2 - 1}{t^2 + t - 2} \), find f′(t).
Answer:
\( f'(t) = \frac{(t^2 + t - 2)(2t) - (t^2 - 1)(2t + 1)}{(t^2 + t - 2)^2} \)
\( = \frac{2t^3 + 2t^2 - 4t - 2t^3 - t^2 + 2t + 1}{(t^2 + t - 2)^2} = \frac{t^2 - 2t + 1}{(t^2 + t - 2)^2} \)

Question 31. \( z = \frac{2x + 1}{x^2 - 1} \)
Answer:
\( \frac{dz}{dx} = \frac{(x^2 - 1)(2) - (2x + 1)(2x)}{(x^2 - 1)^2} = \frac{2x^2 - 2 - 4x^2 - 2x}{(x^2 - 1)^2} = \frac{-2x^2 - 2x - 2}{(x^2 - 1)^2} \)

Question 32. \( y = \frac{\sin x + \cos x}{\cos x} \)
Answer:
\( \frac{dy}{dx} = \frac{\cos x(\cos x - \sin x) - (\sin x + \cos x)(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \sec^2 x \)

Question 33. \( y = \frac{\cot x}{1 + \cot x} \)
Answer:
\( \frac{dy}{dx} = \frac{(1 + \cot x)(-\text{cosec}^2 x) - \cot x(-\text{cosec}^2 x)}{(1 + \cot x)^2} = \frac{-\text{cosec}^2 x}{(1 + \cot x)^2} \)

Question 34. \( y = \frac{\cos x}{x} + \frac{x}{\cos x} \)
Answer:
\( \frac{d}{dx}\left(\frac{\cos x}{x}\right) = \frac{x(-\sin x) - \cos x}{x^2} \),  \( \frac{d}{dx}\left(\frac{x}{\cos x}\right) = \frac{\cos x - x(-\sin x)}{\cos^2 x} \)
\( \therefore \frac{dy}{dx} = \frac{-x\sin x - \cos x}{x^2} + \frac{x\sin x + \cos x}{\cos^2 x} \)

Question 35. \( p = \frac{\tan q}{1 + \tan q} \)
Answer:
\( \frac{dp}{dq} = \frac{(1 + \tan q)\sec^2 q - \tan q\sec^2 q}{(1 + \tan q)^2} = \frac{\sec^2 q}{(1 + \tan q)^2} \)

Differentiation by Chain Rule

Find dy/dx as a function of x.

Question 36. y = (2x + 1)⁵
Answer:
With u = 2x + 1, y = u⁵: \( \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = 5u^4\cdot 2 = 10(2x + 1)^4 \)

Question 37. y = (4 − 3x)⁹
Answer:
\( \frac{dy}{dx} = 9(4 - 3x)^8(-3) = -27(4 - 3x)^8 \)

Question 38. \( y = \left(1 - \frac{x}{7}\right)^{-7} \)
Answer:
With \( u = 1 - \frac{x}{7} \), y = u⁻⁷: \( \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = -7u^{-8}\cdot\left(-\frac{1}{7}\right) = \left(1 - \frac{x}{7}\right)^{-8} \)

Question 39. \( y = \left(\frac{x}{2} - 1\right)^{-10} \)
Answer:
\( \frac{dy}{dx} = -10\left(\frac{x}{2} - 1\right)^{-11}\cdot\frac{1}{2} = -5\left(\frac{x}{2} - 1\right)^{-11} \)

Question 40. y = sin 5x
Answer:
\( \frac{dy}{dx} = \cos 5x\cdot 5 = 5\cos 5x \)

Question 41. y = sin(x) + ln(x²) + e^2x
Answer:
\( \frac{dy}{dx} = \cos x + \frac{1}{x^2}\cdot 2x + e^{2x}\cdot 2 = \cos x + \frac{2}{x} + 2e^{2x} \)

Question 42. y = 2 sin(ωx + φ), where ω and φ are constants
Answer:
\( \frac{dy}{dx} = 2\cos(\omega x + \phi)\cdot\omega = 2\omega\cos(\omega x + \phi) \)

Question 43. y = sin³ x + sin 3x
Answer:
\( \frac{dy}{dx} = 3\sin^2 x\cos x + 3\cos 3x \)

Question 44. y = sin²(x² + 1)
Answer:
\( \frac{dy}{dx} = 2\sin(x^2 + 1)\cdot\cos(x^2 + 1)\cdot 2x = 4x\sin(x^2 + 1)\cos(x^2 + 1) \)

Question 45. y = x(x² + 1)^−1/2
Answer:
\( \frac{dy}{dx} = (x^2 + 1)^{-1/2} + x\cdot\left(-\frac{1}{2}\right)(x^2 + 1)^{-3/2}(2x) = \frac{(x^2 + 1) - x^2}{(x^2 + 1)^{3/2}} \)
\( = \frac{1}{(x^2 + 1)^{3/2}} \)

Question 46. \( q = \sqrt{2r - r^2} \), find \( \frac{dq}{dr} \).
Answer:
\( \frac{dq}{dr} = \frac{1}{2\sqrt{2r - r^2}}\cdot(2 - 2r) = \frac{1 - r}{\sqrt{2r - r^2}} \)

Question 47. \( y = \left(\frac{x^2}{8} + x - \frac{1}{x}\right)^4 \)
Answer:
With \( u = \frac{x^2}{8} + x - \frac{1}{x} \), y = u⁴:
\( \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = 4u^3\left(\frac{x}{4} + 1 + \frac{1}{x^2}\right) = 4\left(\frac{x^2}{8} + x - \frac{1}{x}\right)^3\left(\frac{x}{4} + 1 + \frac{1}{x^2}\right) \)

Given y = f(u) and u = g(x), Find dy/dx

Question 48. y = 2u³, u = 8x − 1
Answer:
\( \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = 6u^2\cdot 8 = 48(8x - 1)^2 \)

Question 49. y = sin u, u = 3x + 1
Answer:
\( \frac{dy}{dx} = \cos u\cdot 3 = 3\cos(3x + 1) \)

Question 50. y = 6u − 9, u = (1/2)x⁴
Answer:
\( \frac{dy}{dx} = 6\cdot 2x^3 = 12x^3 \)

Question 51. y = cos u, \( u = -\frac{x}{3} \)
Answer:
\( \frac{dy}{dx} = -\sin u\cdot\left(-\frac{1}{3}\right) = \frac{1}{3}\sin\left(-\frac{x}{3}\right) = -\frac{1}{3}\sin\frac{x}{3} \)

Question 52. y = cos u, u = sin x
Answer:
\( \frac{dy}{dx} = -\sin u\cdot\cos x = -\sin(\sin x)\cos x \)

Question 53. y = sin u, u = x − cos x
Answer:
\( \frac{dy}{dx} = \cos u\cdot(1 + \sin x) = \cos(x - \cos x)(1 + \sin x) \)

Second Derivatives

Question 54. Find y″ if (a) y = cos x (b) y = sec x
Answer:
(a) y′ = −sin x, y″ = −cos x
(b) y′ = sec x tan x
y″ = sec x tan x · tan x + sec x · sec² x = sec x tan² x + sec³ x = sec x(sec² x − 1) + sec³ x = 2 sec³ x − sec x

Question 55. Position of a particle is given by S = t³ − 2t² + 5t + 4. (a) Find the position of the particle at t = 1 s. (b) Find the first derivative of S at t = 1 s. (c) Find the second derivative of S at t = 1 s.
Answer:
(a) S(1) = 1 − 2 + 5 + 4 = 8
(b) \( \frac{dS}{dt} = 3t^2 - 4t + 5 \); at t = 1: 3 − 4 + 5 = 4
(c) \( \frac{d^2S}{dt^2} = 6t - 4 \); at t = 1: 6 − 4 = 2

Differentiation as a Rate Measurement

Question 56. Suppose that the radius r and area A = πr² of a circle are differentiable functions of t. Write an equation that relates dA/dt to dr/dt.
Answer:
\( \frac{dA}{dt} = \frac{dA}{dr}\cdot\frac{dr}{dt} = 2\pi r\frac{dr}{dt} \)

Question 57. Suppose that the radius r and surface area S = 4πr² of a sphere are differentiable functions of t. Write an equation that relates \( \frac{dS}{dt} \) to \( \frac{dr}{dt} \).
Answer:
\( \frac{dS}{dt} = \frac{dS}{dr}\cdot\frac{dr}{dt} = 8\pi r\frac{dr}{dt} \)

Question 58. The radius r and height h of a circular cylinder are related to the cylinder's volume V by the formula V = πr²h.
(a) If the height is increasing at a rate of 5 m/s while the radius is constant, find the rate of increase of the volume of the cylinder.
(b) If the radius is increasing at a rate of 5 m/s while the height is constant, find the rate of increase of the volume of the cylinder.
(c) If the height is increasing at a rate of 5 m/s and the radius is increasing at a rate of 5 m/s, find the rate of increase of the volume of the cylinder.
Answer:
(a) \( \frac{dV}{dt} = \pi r^2\frac{dh}{dt} = 5\pi r^2 \)
(b) \( \frac{dV}{dt} = 2\pi rh\frac{dr}{dt} = 10\pi rh \)
(c) \( \frac{dV}{dt} = \pi r^2\frac{dh}{dt} + 2\pi rh\frac{dr}{dt} = 5\pi r^2 + 10\pi rh \)

Maxima and Minima

Question 59. Find the maximum and minimum values of the function 2x³ − 15x² + 36x + 11.
Answer:
y = 2x³ − 15x² + 36x + 11; \( \frac{dy}{dx} = 6x^2 - 30x + 36 = 6(x - 2)(x - 3) \)
dy/dx = 0 gives x = 2 or x = 3.
\( \frac{d^2y}{dx^2} = 12x - 30 \): at x = 2 it is −6 (< 0, maximum); at x = 3 it is 6 (> 0, minimum).
y_max = 2(8) − 15(4) + 36(2) + 11 = 39;  y_min = 2(27) − 15(9) + 36(3) + 11 = 38

Question 60. Find two positive numbers x and y such that x + y = 60 and xy is maximum.
Answer:
y = 60 − x, so P = xy = 60x − x²; \( \frac{dP}{dx} = 60 - 2x = 0 \Rightarrow x = 30 \)
\( \frac{d^2P}{dx^2} = -2 < 0 \), so P is maximum.
x = 30 and y = 30

Question 61. A sheet of area 40 m² is used to make an open tank with a square base. Find the dimensions of the base such that the volume of this tank is maximum.
Answer:
Let the side of the square base be x and the height be h. Area of sheet (base + 4 sides): x² + 4xh = 40 ⇒ \( h = \frac{40 - x^2}{4x} \)
Volume \( V = x^2h = \frac{x(40 - x^2)}{4} = 10x - \frac{x^3}{4} \)
\( \frac{dV}{dx} = 10 - \frac{3x^2}{4} = 0 \Rightarrow x^2 = \frac{40}{3} \); \( \frac{d^2V}{dx^2} = -\frac{3x}{2} < 0 \), so V is maximum.
Side of the base \( x = \sqrt{\frac{40}{3}} \) m (and then the height is h = x/2).

Question 62. State true or false: If a function has a maximum value at point P, then the slope of the tangent drawn to the function at point P is zero.
Answer:
True. At a maximum (of a smooth curve) dy/dx = 0, so the tangent at P is parallel to the x-axis and its slope is zero.

Integration

Find the integrals of the given functions (C is the constant of integration).

Question 63. (a) 2x (b) x² (c) x² − 2x + 1
Answer:
(a) \( \int 2x\,dx = x^2 + C \)
(b) \( \int x^2\,dx = \frac{x^3}{3} + C \)
(c) \( \int (x^2 - 2x + 1)\,dx = \frac{x^3}{3} - x^2 + x + C \)

Question 64. (a) −3x⁻⁴ (b) x⁻⁴ (c) x⁻⁴ + 2x + 3
Answer:
(a) \( \int -3x^{-4}dx = -3\cdot\frac{x^{-3}}{-3} + C = x^{-3} + C \)
(b) \( \int x^{-4}dx = -\frac{1}{3}x^{-3} + C \)
(c) \( \int (x^{-4} + 2x + 3)\,dx = -\frac{1}{3}x^{-3} + x^2 + 3x + C \)

Question 65. (a) \( \frac{1}{x^2} \) (b) \( \frac{5}{x^2} \) (c) \( 2 - \frac{5}{x^2} \)
Answer:
(a) \( \int x^{-2}dx = -\frac{1}{x} + C \)
(b) \( \int 5x^{-2}dx = -\frac{5}{x} + C \)
(c) \( \int \left(2 - \frac{5}{x^2}\right)dx = 2x + \frac{5}{x} + C \)

Question 66. (a) \( \frac{3}{2}\sqrt{x} \) (b) \( \frac{3}{2\sqrt{x}} \) (c) \( \sqrt{x} + \frac{1}{\sqrt{x}} \)
Answer:
(a) \( \int \frac{3}{2}x^{1/2}dx = \frac{3}{2}\cdot\frac{x^{3/2}}{3/2} + C = \sqrt{x^3} + C \)
(b) \( \int \frac{3}{2}x^{-1/2}dx = \frac{3}{2}\cdot\frac{x^{1/2}}{1/2} + C = 3\sqrt{x} + C \)
(c) \( \int (x^{1/2} + x^{-1/2})\,dx = \frac{2\sqrt{x^3}}{3} + 2\sqrt{x} + C \)

Question 67. (a) \( \frac{4}{3}\sqrt[3]{x} \) (b) \( \frac{1}{3\sqrt[3]{x}} \) (c) \( \sqrt[3]{x} + \frac{1}{\sqrt[3]{x}} \)
Answer:
(a) \( \int \frac{4}{3}x^{1/3}dx = \frac{4}{3}\cdot\frac{x^{4/3}}{4/3} + C = x^{4/3} + C \)
(b) \( \int \frac{1}{3}x^{-1/3}dx = \frac{1}{3}\cdot\frac{x^{2/3}}{2/3} + C = \frac{x^{2/3}}{2} + C \)
(c) \( \int (x^{1/3} + x^{-1/3})\,dx = \frac{3x^{4/3}}{4} + \frac{3x^{2/3}}{2} + C \)

Question 68. (a) \( \frac{1}{2}x^{-1/2} \) (b) \( -\frac{1}{2}x^{-3/2} \) (c) \( -\frac{3}{2}x^{-5/2} \)
Answer:
(a) \( \int \frac{1}{2}x^{-1/2}dx = x^{1/2} + C \)
(b) \( \int -\frac{1}{2}x^{-3/2}dx = x^{-1/2} + C \)
(c) \( \int -\frac{3}{2}x^{-5/2}dx = x^{-3/2} + C \)

Question 69. (1 − x² − 3x⁵)
Answer:
\( \int (1 - x^2 - 3x^5)\,dx = x - \frac{x^3}{3} - \frac{x^6}{2} + C \)

Question 70. 3 sin x
Answer:
\( \int 3\sin x\,dx = -3\cos x + C \)

Question 71. \( \frac{1}{3x} \)
Answer:
\( \int \frac{1}{3x}dx = \frac{1}{3}\ln x + C \)

Question 72. \( \int (2x^3 - 5x + 7)\,dx \)
Answer:
\( = \frac{2x^4}{4} - \frac{5x^2}{2} + 7x + C = \frac{x^4}{2} - \frac{5x^2}{2} + 7x + C \)

Question 73. \( \int \left(\frac{1}{5} - \frac{2}{x^3} + 2x\right)dx \)
Answer:
\( = \frac{x}{5} - 2\cdot\frac{x^{-2}}{-2} + x^2 + C = \frac{x}{5} + \frac{1}{x^2} + x^2 + C \)

Question 74. \( \int \left(\sqrt{x} + \sqrt[3]{x}\right)dx \)
Answer:
\( = \frac{x^{3/2}}{3/2} + \frac{x^{4/3}}{4/3} + C = \frac{2}{3}x^{3/2} + \frac{3}{4}x^{4/3} + C \)

Question 75. \( \int x^{-3}(x + 1)\,dx \)
Answer:
\( = \int (x^{-2} + x^{-3})\,dx = -\frac{1}{x} - \frac{1}{2x^2} + C \)

Question 76. \( \int \frac{t\sqrt{t} + \sqrt{t}}{t^2}dt \)
Answer:
\( = \int (t^{-1/2} + t^{-3/2})\,dt = 2\sqrt{t} - \frac{2}{\sqrt{t}} + C \)

Question 77. \( \int \frac{4 + \sqrt{t}}{t^3}dt \)
Answer:
\( = \int (4t^{-3} + t^{-5/2})\,dt = -2t^{-2} - \frac{2}{3}t^{-3/2} + C \)

Question 78. \( \int \cos\theta(\tan\theta + \sec\theta)\,d\theta \)
Answer:
\( = \int (\sin\theta + 1)\,d\theta = -\cos\theta + \theta + C \)

Integrate by using the substitution suggested in the bracket.

Question 79. \( \int \sin 3x\,dx \) (use u = 3x)
Answer:
u = 3x, du = 3 dx: \( \int \sin 3x\,dx = \frac{1}{3}\int \sin u\,du = -\frac{1}{3}\cos u + C = -\frac{1}{3}\cos 3x + C \)

Question 80. \( \int \sec 2t\tan 2t\,dt \) (use u = 2t)
Answer:
u = 2t, du = 2 dt: \( \int \sec 2t\tan 2t\,dt = \frac{1}{2}\int \sec u\tan u\,du = \frac{1}{2}\sec 2t + C \)

Definite Integrals and Area

Question 81. \( \int_{-4}^{-1} \frac{\pi}{2}d\theta \)
Answer:
\( = \frac{\pi}{2}[\theta]_{-4}^{-1} = \frac{\pi}{2}(-1 + 4) = \frac{3\pi}{2} \)

Question 82. \( \int_0^1 e^x\,dx \)
Answer:
\( = [e^x]_0^1 = e - 1 \)

Question 83. \( \int_{\pi}^{2\pi} \theta\,d\theta \)
Answer:
\( = \left[\frac{\theta^2}{2}\right]_{\pi}^{2\pi} = \frac{4\pi^2 - \pi^2}{2} = \frac{3\pi^2}{2} \)

Question 84. \( \int_0^{\sqrt[3]{7}} x^2\,dx \)
Answer:
\( = \left[\frac{x^3}{3}\right]_0^{\sqrt[3]{7}} = \frac{7}{3} \)

Question 85. \( \int_0^1 \frac{dx}{3x + 2} \)
Answer:
\( = \left[\frac{1}{3}\ln(3x + 2)\right]_0^1 = \frac{1}{3}(\ln 5 - \ln 2) = \frac{1}{3}\ln\frac{5}{2} \)

Use a definite integral to find the area of the region between the given curve and the x-axis on the interval [0, b].

Question 86. y = 2x
Answer:
\( \text{Area} = \int_0^b 2x\,dx = [x^2]_0^b = b^2 \)

Question 87. \( y = \frac{x}{2} + 1 \)
Answer:
\( \text{Area} = \int_0^b \left(\frac{x}{2} + 1\right)dx = \left[\frac{x^2}{4} + x\right]_0^b = \frac{b^2}{4} + b = \frac{b(4 + b)}{4} \)

Question 88. y = 3x²
Answer:
\( \text{Area} = \int_0^b 3x^2\,dx = [x^3]_0^b = b^3 \)

Question 89. \( y = \sqrt{b^2 - x^2} \)
Answer:
The curve is the upper half of the circle x² + y² = b². On [0, b] the region is a quarter of a circle of radius b.
\( \text{Area} = \int_0^b \sqrt{b^2 - x^2}\,dx = \frac{1}{4}\pi b^2 = \frac{\pi b^2}{4} \)

Objective Questions