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

Asked on by fakeyelid

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

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