Find d y d x for the following y = x 7 2 .

Given function is y = x 7 2 We have to find d y d x for the given function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables. Now, y = x 7 2 Differentiate with respect to the x . d y d x = 7 2 x 7 2 - 1 d y d x = 7 2 x 5 2

Find d y d x for the following y = x – 3

given y = x – 3 we have to find differentiation of y with respect to x d d x ( x - 3 ) = - 3 x - 3 - 1 ∴ d d x x n = n . x n - 1 d y d x = - 3 x - 4

Evaluate each of the following integrals: ∫ a x 2 + b x + c d x

Given expressions: ∫ a x 2 + b x + c d x As we, know that ∫ x n d x = x n + 1 n + 1 + C Comparing above expressions, then = ∫ a x 2 + b x + c d x = ∫ a x 2 d x + ∫ b x d x + ∫ c d x = a x 2 + 1 2 + 1 + b x 1 + 1 1 + 1 + c x + C = a x 3 3 + b x 2 2 + c x + C = a x 3 3 + b x 2 2 + c x + C

Evaluate the following integrals ∫ r 1 r 2 – k q 1 q 2 x 2 d x

Given, function is ∫ r 1 r 2 – k q 1 q 2 x 2 d x A definite integral is one that has lower and upper limits and on solving gives a constant result. Now, ∫ r 1 r 2 – k q 1 q 2 x 2 d x = – k q 1 q 2 ∫ r 1 r 2 x - 2 d x = – k q 1 q 2 x - 2 + 1 - 2 + 1 r 1 r 2 = k q 1 q 2 x - 1 r 1 r 2 = k q 1 q 2 1 x r 1 r 2 = k q 1 q 2 1 r 2 – 1 r 1

Evaluate the following integrals ∫ u v M v d v

Given, function ∫ u v M v d v A definite integral is one that has lower and upper limits and on solving gives a constant result. Now, ∫ u v M v d v Use formula, ∫ x n d x = x n + 1 n + 1 = M ∫ u v v d v = M v 2 2 u v = 1 2 M v 2 u v = 1 2 M v 2 - u 2

Evaluate the following integrals ∫ 0 ∞ x 1 2 d x

Given function, ∫ 0 ∞ x 1 2 dx A definite integral is one that has lower and upper limits and on solving gives a constant result. Now, ∫ 0 ∞ x 1 2 d x Use formula, ∫ x n d x = x n + 1 n + 1 + C ∴ = x 1 2 + 1 1 2 + 1 0 ∞ = 2 3 x 3 2 0 ∞ = 2 3 ∞ - 0 3 2 = 2 3 × ∞ = ∞