Calculus Questions

This document contains solved calculus questions and answers for applied mathematics. It covers topics such as differential equations, integration, and area under the curve problems. The solutions are provided step-by-step for better understanding. This study material is suitable for students studying calculus.

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Calculus Questions
Applied Mathematics

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Calculus Questions
Task-1
a.i. Given : y = 4Asinx + 2Bcosx
Differentiating wrt x-
y’= 4Acosx – 2Bsinx
Again Differentiating wrt x-
y’’= - 4Asinx - 2Bcosx
y’’= -y or
y’’+ y = 0
Since we differentiate it twice, it is 2nd order differential equation
a.ii. Given : y = x + A
x
Differentiating wrt x-
y’= 1 - A
x2
y’= 1 - 1
x (y – x)
y’+ y
x - 1 = 1
y’+ y
x = 2
It is 1st Order Differential Equation
a.iii. Given : y = Ax2 – Bx (1)
Differentiating wrt x-
y’= 2Ax – B (2)
Again Differentiating wrt x-
y’’= 2A or A = y’’/2 (3)
Put (3) in (2)-

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Calculus Questions
y’= y’’x – B or B = y’’x – y’ (4)
Put (3) and (4) in (1)-
y = y ' '
2 x2 – y’’x2 – y’x
y ' '
2 x2 + y’x + y = 0
It is a 2nd Order Differential Equation.
b.i. Given : dy
dx = y
x
dy
y = dx
x
Integrating both sides-
∫ dy
y =¿∫ dx
x ¿
ln(y) = ln(x) + ln(c) where ln(c) is the constant of integration
ln(y) = ln(xc) or
y = xc is the general solution of the given differential equation.
b.ii. Given : cosx
3+ y
dy
dx = sinx
Separating the variables-
dy
3+ y = tanx dx
Integrating both sides-
∫ dy
3+ y =∫tanx dx
ln(3 + y) = ln(secx) + ln(c) where ln(c) is the constant of integration
ln(3 + y) = ln(c*secx)
3 + y = c*secx

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Calculus Questions
y = c*secx – 3 is the general solution of the given differential
equation.
b.iii. Given : dy
dx = (2y + 1)(x – 3)
Separating the variables-
dy
2 y +1 = (x – 3)dx
Integrating both sides-
∫ dy
2 y+1=∫( x – 3)dx
1
2 ln(2y + 1) = x2
2 – 3x + C
ln(2y + 1) = x2 – 6x + 2C
2y + 1 = ex2−6 x+2 C
y = 1
2 (e x2−6 x+2 C )− 1
2
c.i. Given : -3 d2 y
d x2 + 36 dy
dx - 108y = 0
Putting d
dx ≡ D∧d2
d x2 ≡ D2
Then, (-3D2 + 36D – 108)y = 0
Solving the above auxiliary equation which is quadratic.
D = −b ± √b2−4 ac
2a = −36 ± √1296−1296
−6 = 6 or 6
Solution of the differential equation is : y = (A + Bx)e6x
c.ii. Given : d2 y
d x2 + 6 dy
dx + 9y = 0

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Calculus Questions
Putting d
dx ≡ D∧d2
d x2 ≡ D2
Then, (D2 + 6D + 9)y = 0
Solving the above auxiliary equation which is quadratic1
D = −b ± √b2−4 ac
2a = −6 ± √36−36
2 = -3 or -3
Solution of the differential equation is : y = (A + Bx)e-3x
Task-2
Let the length of the rectangular window be a, breadth be b and radius of the
semicircle surmounted on rectangle be r, then, the shape would be like-
b
a
Here, r = a/2
According to question,
2b + a + πa
2 = 10 (a and b are in metre)
b = 5 – a
2 - πa
4 (1)
Now, area of the window will be the sum of area of rectangle and semicircle i.e.,
A = ab + π a2
4

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Calculus Questions
Putting equation (1)-
A = a(5 – a
2 - πa
4 ) + π a2
4
A = 5a - a2
2
For Area to be maximum, we have to differentiate wrt a,
dA
dx = 5 – a = 0 (For Maxima or Minima)
a = 5
Also, d2 A
d x2 = - 1
So, a = 5 is a point of Maxima for Area.
Maximum Amount of light pass through the window when length of the rectangle is
5m, breadth of the rectangle is ( 5
2 - 5 π
4 )m and radius of the semicircle is 2.5m
Task-3
i. y = ∫ e−3 x cos 2 xdx (1)
Using Integration by parts, considering e-3x as the 1st function and cos2x as the 2nd
function.
y = e-3x ∫cos 2 xdx + 3
2 ∫e−3 x sin 2 x dx
y = e−3 x sin2 x
2 + 3
2( e-3x∫sin 2 xdx - 3
2 ∫e−3 x cos 2 xdx )
y = e−3 x sin2 x
2 + 3 e−3 x cos 2 x
4 - 9
4 y (From (1))
13
4 y = e−3 x sin2 x
2 + 3 e−3 x cos 2 x
4
y = 4
13 ( e−3 x sin2 x
2 + 3 e−3 x cos 2 x
4 )
ii. Given : y = ∫ 2 x2+ x +1
( x−1 ) ( x2+1 ) dx

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Calculus Questions
Separating Partial Fractions2-
Let 2 x2 + x +1
( x−1 ) ( x2 +1 ) = A
( x−1 ) + Bx +C
( x2+1 )
2x2 + x + 1 = (A + B)x2 + (-B + C)x + (A-C)
Comparing the coefficients-
A = 2, B = 0, C = 1
So, 2 x2 + x +1
( x−1 ) ( x2 +1 ) = 2
( x−1 ) + 1
( x2+1 )
Now,
y = ∫ ( 2
( x−1 ) + 1
( x2+ 1 ) )dx
= 2ln(x – 1) + tan-1x + C
Task-4
i. Function of the curve : y = 5ex
Area under the curve = ∫ ydx
Area is to be calculated from x = 0 to x = 3
Area = ∫
0
3
5 e x
= 5[e3 – e0]
= 5[20.085 – 1]
= 95.43 sq.unit
ii. Given : Equation of straight line is y = -2x + 28
And Equation of curve is y = 36 – x2
For finding the points of intersection of the line and the curve, we have to
equate both equations-
36 – x2 = -2x + 28
x2 – 2x – 8 = 0
(x + 2)(x – 4) = 0
Therefore, x = -2 or 4 are the points at which both the line and the curve
intersect.

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Calculus Questions
Area enclosed by the curve = ∫
−2
4
ycurve dx−∫
−2
4
yline dx
= ∫
−2
4
( 36−x2 ) dx−∫
−2
4
( −2 x+28 ) dx
= 36 (4 + 2) – 64+8
3 + (16 – 4) + 28 (4 + 2)
= 216 – 24 + 12 + 168
= 12 sq. unit
References-
1Sharma R.D. (2018). Mathematics (2019 edition). Dhanpat Rai Publication, New Delhi
2Aggarwal, R.S. (2013). Senior Secondary School Mathematics (2018 edition). Bharti
Bhawan Publishers, New Delhi