# Evaluate the definite integral of y=4x^7+5x^-1+6sinx if x=1 to x=2.

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We have to find the definite integral of y = 4x^7 + 5x^-1 + 6sinx if x=1 to x=2.

Int [ y] = Int [4x^7 + 5x^-1 + 6 sin x]

=> Int [4x^7] + Int [5x^-1] + Int [6 sin x]

=> (4/8)x^8 + 5 ln x - 6 cos x

Now for x = 2

(4/8)x^8 + 5 ln x - 6 cos x

= (4/8) 2^8 + 5 ln 2 - 6 cos 2

= 128 + 5 ln 2 - 6 cos 2

For x = 1

(4/8)x^8 + 5 ln x - 6 cos x

= (4/8) - 6 cos x

128 + 5 ln 2 - 6 cos 2 - (4/8) + 6 cos 1

=> 127.5 + 5 ln 2 - 6( cos 2 - cos 1)

**The required result is 127.5 + 5 ln 2 - 6( cos 2 - cos 1)**

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

First, we'll determine the indefinite integral of f(x).

Int f(x)dx = Int (4x^7+5x^-1+6sinx)dx

We'll apply the additive property of integrals:

Int (4x^7+5x^-1+6sinx)dx = Int 4x^7dx + Int 5x^-1dx + Int6sinxdx

We'll re-write the integrals:

Int f(x)dx = 4Int x^7dx + 5Int dx/x + 6Int sinxdx

Int f(x)dx = 4x^8/8 + 5ln (x) - 6 cos x + C

The definite integral is:

Int f(x)dx = F(b) - F(a)

for x = 1

F(2) = 2^8/2 + 5ln (2) - 6 cos 2

F(2) = 2^7 + ln (2^5) - 6 cos 2

F(2) = 128 + ln (2^5) - 6 cos 2

F(1) = 1^8/2 + 5ln (1) - 6 cos 1

F(1) = 1/2 - 6 cos 1

F(2) - F(1) = 128 + ln (2^5) - 6 cos 2 - 1/2 + 6 cos 1

F(2) - F(1) = 127.5 + ln 32 - 6(cos 2 - cos 1)

**Int f(x)dx = 127.5 + ln 32 - 6(cos 2 - cos 1)**