Introduction of calculus :
Calculus mainly used to evaluate the definite integrals and differentiation. Main objective of the calculus is to obtain the rate of change and motion. Calculus has classified into two major branches,
1) Differential calculus
2) Integral calculus
In differential calculus, we have to differentiate the given equations and reduce the losses in real life. In Integral calculus, we have use the integration process and maximize the values in real life.
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Ex : 1
If y = 4x3 + 2x2
(i) Differentiate y.
(ii) Find the value of dy when x = 3 and dx = 7
(i) Given y = 4x3 + 2x2,
Differentiate y with respect to x,
dy= 12x2 dx + 4x dx
dy = (12x2 + 4x) dx
(ii) Substituting x = 3 and dx = 7
We get dy = (12 × 32 + 4 × 3) 7 = 840.
The final answer is 840.
Ex : 2
The luminous intensity I candelas of a lamp at varying voltage V is given by : I = 6 × 10?4V2. Determine the voltage at which the light is increasing at a rate of 1.2 candelas per volt.
The rate of change of light with respect to voltage is given by (dI/dV).
Since I = 6 × 10?4V2
(dI/dV) = 1.2 × 10?4V.
When the light is increasing at 1.2 candelas per volt then
(dI/dV) = + 1.2.
We must have + 1.2 = 12 × 10-4 V, from which,
Voltage V = 1.2 / (12 × 10?4) = 0.1 × 104 = 1000 Volts.
The final answer is 1000 Volts.
Ex : 3
? x12 dx
? x12 dx = (x(12 + 1)/ 12 + 1) + c.
= (x13/13) + c.
? x12 dx = (x13/13) + c.
Ex : 4
? 3x2 dx
? 3x2 dx = 3 (x(2 + 1)/ 2 +1) + c.
= 3 (x3/3) + c.
= x3 + c.
? 3x2 dx = x3 + c.
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|Vector Calculus||Maxima and Minima|
|Mean Value Theorem||Limits|
|How to Find The Constant of Integration|