What is the integral from (11pi)/2 to 6pi of 9sin(x) ?

The integral from (11pi)/2 to 6pi of 9sin(x) is -9

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integral from (11pi)/2 to 6pi of 9sin(x)

\int_{\frac{11 π}{2}}^{6 π} 9 sin (x) d x

- 9

Solution steps

\int_{\frac{11 π}{2}}^{6 π} 9 sin (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int_{\frac{11 π}{2}}^{6 π} sin (x) d x

Use the common integral:

\int sin (x) d x = - cos (x)

= 9 [- cos (x)]_{\frac{11 π}{2}}^{6 π}

Compute the boundaries:

- 1

= 9 (- 1)

Simplify

= - 9

y^{^{'}} = (1 - y) cos (x)

\int \frac{y}{x} d x

y = (x^{3} - 16 x)^{8}, (- 4, 0)

\frac{d y}{d x} = (x + y)

\frac{d}{d x} (x (ln (x^{2})))

What is the integral from (11pi)/2 to 6pi of 9sin(x) ?
The integral from (11pi)/2 to 6pi of 9sin(x) is -9