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

### 2 Answers | Add Yours

The function given is f(x) = 3*x^6 + 3*x^(-1) - cosx. We need the definite integral of f(x) between the values of x = 0 and x = 1.

F(x) = Int [ f(x)]

=> Int [ 3*x^6 + 3*x^(-1) - cos x dx]

=> Int [ 3*x^6 dx]+ Int[ 3*x^(-1) dx] - Int[ cos x dx]

=> (3/7)*x^7 + 3*ln x - sin x + C

F(1) = (3/7) + 0 - sin 1 + C

F(0) = 0 + ln 0 - sin 0 + C

Here we see that we have ln 0, which is not defined. This does not allow us to find the definite integral.

**The definite integral cannot be found as some of the terms for the given limits are not defined.**

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.**