What is the definite integral of y=3x^6+3x^-1-cosx if the limits of integration are 0 and 1.

To determine the value of definite integral, we'll apply
the fundamental theorem of calculus.

We'll determine the
indefinite integral of y.

Int ydx = Int (3x^6 + 3x^-1 -
cosx)dx

We'll apply the additive property of
integrals:

Int (3x^6 + 3x^-1 - cosx)dx = Int 3x^6dx + Int
3x^-1dx - Int cosx dx

We'll re-write the
integrals:

Int ydx = 3Int x^6dx + 3Int dx/x - Int cos x
dx

Int ydx = 3x^7/7 + 3ln (x) -  sin x +
C

According to Leibniz Newton identity, the definite
integral is:

Int ydx = F(b) -
F(a)

F(1) = 3/7 + 3ln (1) - sin
1

F(1) = 3/7 - sin 1

F(0) =
3*0^7/7 + 3ln (0) -  sin 0 + C

Since the term (ln 0) is
undefined, we cannot continue to determine the value of definite
integral.

The definite integral of the
function y=3x^6+3x^-1-cosx doesn't exist, when the limits of integration are x = 0 to x
= 1.