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Integration
We have, so far, seen that differential calculus measures the rate of change of functions. Differentiation is the process of finding the derivative (rate of change) of a function F(x) and is denoted by F'(x) Often, we may know the rate of change, F'(x) of a function F(x) which is unknown to us. In such situations we would like to find out the original function F(x) from the derivative, F'(x). Reversing the process of differentiation and finding out the original function from the derivative is called integration. The original function, F(x) is called the integral.
The left hand side of the equation is read as "the indefinite integral of f(x) with respect to x. The symbol is the integral sign, f(x) is the integrand and 'c' is an arbitrary constant. The arbitrary constant 'c' is added because of the following reason:
If d/dx {F(x)} = f(x) then we can also write that d/dx {F(x) + c} = f(x) where 'c' is an arbitrary constant, because the derivative of any constant is zero.
All the number sets we have seen above put together comprise the real numbers. Real numbers are also inadequate in the sense that it does not include a quantity which i
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from 0->1: Int sqrt(1-x^2) Solution) I=∫sqrt(1-x 2 )dx = sqrt(1-x 2 )∫dx - ∫{(-2x)/2sqrt(1-x 2 )}∫dx ---->(INTEGRATION BY PARTS) = x√(1-x 2 ) - ∫-x 2 /√(1-x 2 ) Let
Using the formulas and properties from above find out the value of the subsequent summation. c The first thing that we require to do here is square out the stuff being summe
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Evaluate given integrals. ∫3/(5 y + 4) dy Solution Let's notice as well that if we take the denominator & differentiate it we get only a constant and th
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