Some basic formulas used in solving this exercise are added below:
|
Derivatives |
Integrals (Anti-Derivatives) |
|---|---|
|
d/dx (xn+1/ n+1) = xn |
∫ xn dx = xn+1 / (n+1) + C |
|
d(sinx)/ dx = cosx |
∫dx = x + C |
|
d(cosx)/ dx = -sinx |
∫ cosx dx = sinx + C |
|
d(tanx)/ dx = -sec2x |
∫ sinx dx = -cosx + C |
|
d(-cotx)/ dx = cosec2x |
∫ sec2x dx = tanx + C |
|
d(secx)/ dx = secx.tanx |
∫ cosec2x dx = -cotx + C |
|
d(cosecx)/ dx = – cosecx cotx |
∫ secx.tanx dx = secx + C |
|
d(sin-1x)/ dx = 1/√(1-x2) |
∫ cosecx cotx dx = -cosecx + C |
|
d(cos-1x)/ dx = -1/√(1-x2) |
∫ dx/√(1-x2) = sin-1x + C |
|
d(tan-1x)/ dx = 1/(1+x2) |
∫ dx/ (1+x2) = tanx + C |
|
d(ex) / dx = ex |
∫ex dx = ex + C |
|
d(logx )/ dx = 1/x |
∫1/x dx = log|x| + C |
|
d/dx (ax/loga) = ax |
∫ax dx = ax / loga + C |
Chapter 7– Integrals Exercise 7.1
Find an antiderivative (or integral) of the following functions by the method of inspection in Exercises 1 to 5.
Question 1: sin2x
Solution:
Anti-derivative of sin 2x is a function of x whose derivative is sin 2x. It is known that,
d/dx (cos 2x) = – 2sin 2x
sin 2x = – 1/2 d/dx (cos 2x)
sin2x = d/dx (- 1/2 × cos 2x)
Therefore, anti-derivative sin2x is – 1/2 ×cos 2x.
Question 2: cos3x
Solution:
Anti-derivative of cos 3x is a function of x whose derivative is cos 3x. It is known that,
d/dx (sin 3x) = 3cos 3x
or cos3x = 1/3 × d/dx (sin 3x)
or cos 3x = 1/3×d/dx (sin 3x)
Therefore, anti-derivative of cos3x is 1/3 ×sin 3x.
Question 3: e2x
Solution:
Anti-derivative of e2xis a function of x whose derivative is e2x. It is known that,
d(e2x )/dx = 2e2x
e2x = 1/2 × d(e2x )/dx
e2x = d(1/2e2x )/dx
Therefore, anti-derivative of e2x is 1/2 × e2x
Question 4: (ax +b)2
Solution:
Anti-derivative of It is known that (ax + b) 2 is the function of x whose derivative is (ax + b)2
d/dx ((ax + b) 2) = 2a × (ax + b)
(ax + b)22 = 1/(2a) × d/dx ((ax + b) 2)
(ax + b)2 = d/dx (1/(3a) × (ax + b)3)
Therefore, anti-derivative of (ax + b)3 × is 1/(3a) (ax + b)3
Question 5: sin2x -4e3x
Solution:
Anti-derivative of It is known that sin2x -4e3x is the function of x whose derivative is sin2x- 4e3x
We know that d(cos2x)/dx = -2 sin2x
dividing by -2, d(-1/2 cos2x)/ dx = sin2x…(i)
Again d(e3x )/dx = 3e3x
or d(1/3 × e3x )/ dx = e3x
Multiplying by -4, d(-4/3 × e3x )/dx = -4e3x…(ii)
d/ dx (-1/2 cos2x ) + d/dx (-4/3 e3x ) = sin2x – 4e3x
or d/ dx (-1/2 cos2x + -4/3 e3x) = sin2x – 4e3x
Therefore, anti-derivative of sin2x -4e3x is -1/2 cos2x -4/3 e3x
Find the following integrals in Exercise 6 to 20
Question 6: ∫ (4e3x + 1) dx
Solution:
∫ (4e3x + 1) dx
= 4∫e3x dx + ∫ 1dx
= 4 {e3x/3} + x + C
= 4/3 e3x + x + C
Question 7: ∫x2 (1 – 1/x2 ) dx
Solution:
∫x2 (1-1/x2 ) dx
= ∫(x2 – 1 ) dx
= ∫x2 dx – ∫ 1 dx
= x3 / 3 – x + C
Question 8: ∫(ax2 + bx +c ) dx
Solution:
∫(ax2 + bx + c) dx
= a ∫ x2 dx + b ∫x dx + c∫ 1.dx
=a❲x3/3❳ +b❲x2/2❳ + cx + C
= ax3/3 + bx2/2 + cx + C
Question 9: ∫(2x2 + ex) dx
Solution:
∫(2x2 + ex) dx
= 2 ∫x2.dx + ∫ex dx
= 2❲x3/3❳ + ex +C
= 2/3x3 + ex + C
Question 10: ∫ {√x – 1/√x}2 dx
Solution:
∫ {√x – 1/√x}2 dx
opening square
= ∫ {x + 1/x -2} dx
= ∫x dx + ∫ 1/x dx – 2∫1.dx
= x2/2 + log|x| – 2x + C
Question 11: ∫ (x3 + 5x2 -4)/x2 dx
Solution:
∫ (x3 + 5x2 -4) / x2 dx
= ∫(x + 5 -4x-2) dx
= x2/2 + 5x – 4(x-1/-1) + C
= x2 / 2 + 5x +4 /x + C
Question 12: ∫ (x3 + 3x+4)/√x dx
Solution:
∫ (x3 + 3x+4)/√x dx
= ∫ (x5/2 + 3x1/2+4-1/2) dx
= x7/2 /7/2 + 3(x3/2)/ 3/2 + 4(x1/2)/1/2 + C
= 2/7 x7/2 + 2×3/2 + 8x1/2 + C
= 2/7 x7/2 + 2x3/2 8√x + C
Question 13: ∫ (x3 – x2 + x -1 )/ (x-1) dx
Solution:
∫ (x3 – x2 + x -1 )/ (x-1) dx
On dividing (x-1) dx
= ∫ (x2 + 1) dx
= ∫x2dx + ∫ 1 dx
= x3 /3 + x + C
Question 14: ∫ (1 – x) √x dx
Solution:
= ∫ (√x – x3/2) dx
= ∫ x1/2 dx – ∫ x3/2 dx
= x3/2 /3/2 – x5/2 /5/2 + C
= 2/3 x3/2 – 2/3 x5/2 + C
Question 15: ∫ √x (3x2 + 2x + 3) dx
Solution:
∫ √x (3x2+ 2x + 3) dx
= ∫ ❲3x5/2 + 2x3/2 +3x1/2❳dx
= 3∫x5/2 dx + 2∫x3/2dx + 3∫x1/2dx
= 3 ❲x7/2 /7/2❳ + 2❲x5/2 /5/2❳ + 3❲x3/2 /3/2❳ + C
= 6x7/2/7 + 4x5/2/5 + 2x3/2 + C
Question 16: ∫ (2x -3cosx + ex ) dx
Solution:
∫ (2x -3cosx + ex ) dx
= 2 ∫ x dx – 3 ∫cosx dx + ∫ex dx
= 2x/2 – 3(sinx) + ex + C
= x2 – 3sinx + ex + C
Question 17: ∫(2x2-3sinx + 5√x ) dx
Solution:
∫(2x2-3sinx + 5√x) dx
= 2 ∫(x2 dx – ∫3sinx dx + ∫5√x dx
= 2x3 – 3(-cosx) + 5 {x3/2/3/2} + C
= 2/3 x3 + 3cosx + 10/3 x3/2 + C
Question 18: ∫ secx(secx + tanx)dx
Solution:
∫secx(secx + tanx)dx
= ∫(sec2x + secx tanx)dx
= ∫sec2x dx + ∫secx tanx dx
= tanx + secx + C
Question 19: ∫ sec2x /cosec2x dx
Solution:
∫ sec2x /cosec2x dx
= ∫ (1/cos2 x )/(1/sin2x) dx
= ∫ sin2x/ cos2 x dx
= ∫ tan2x dx
= ∫ (sec2x -1 )dx
= ∫ sec2x dx – ∫1 dx
= tanx – x + C
Question 20: ∫ (2-3sinx) / cos2x dx
Solution:
∫ (2 – 3sinx) / cos2x dx
= ∫ (2/cos2x – 3sinx/cosx ) dx
= ∫ 2 sec2x dx – 3∫ tanx secx dx
= 2tanx – 3 secx + C
Question 21: The anti-derivation of (√x + 1/√x) equals
(A) 1/3 x1/3 + 2x1/2 + C
(B) 2/3 x2/3 + 1/2 x2 + C
(C) 2/3 x3/2 + 2x1/2 + C
(D) 3/2 x3/2 + 1/2 x1/2 + C
Correct answer is C
Solution:
(√x + 1/√x) dx
= ∫x1/2 dx + ∫x-1/2 dx
= x3/2 / 3/2+ x1/2 /1/2 + C
= 2/3 x3/2 + 2x1/2 + C
Hence, the correct answer is C.
Question 22: If d/dx f(x) = 4x3 – 3/x4 such that f(2) = 0, then f(x) is
(A) x4 + 1/x3 – 129/8
(B) x3 + 1/x4 + 129/8
(C) x4 + 1/x3 + 129/8
(D) x3 + 1/x4 – 129/8
Correct answer is A
Solution:
f(x) = ∫(4x3 – 3/x4) dx
f(x) = 4∫x3dx – 3∫x-4 dx
f(x) = 4 (x4 /4) – 3(x-3 / -3) + C
f(x) = x4 + 1/x3 + C
Also
f(2) = 0
f(2) = (2)4 + 1/(2)3 + C = 0
C = -(16 + 1/8 ) = -129/8
Therefore f(x) = x4 + 1/x3 – 129/8