What is the integral from 0 to 1 of sin(x)(1-x) ?

The integral from 0 to 1 of sin(x)(1-x) is -sin(1)+1

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integral from 0 to 1 of sin(x)(1-x)

\int_{0}^{1} sin (x) (1 - x) d x

- sin (1) + 1

Decimal Notation

0.15852 \dots

Solution steps

\int_{0}^{1} sin (x) (1 - x) d x

Apply Integration By Parts

: [- cos (x) (1 - x) - \int cos (x) d x]_{0}^{1}

= [- cos (x) (1 - x) - \int cos (x) d x]_{0}^{1}

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

= [- cos (x) (1 - x) - sin (x)]_{0}^{1}

Compute the boundaries:

- sin (1) + 1

= - sin (1) + 1

\int_{0}^{2} (3 x^{2} + 9 - 9 x) d x

\int_{5}^{7} \frac{3}{( x - 6 ) ^{\frac{4}{5}}} d x

\int_{0}^{1} x^{2} + 1 x d x

\int_{- 2 x}^{2 x} 3 x + y d y

\int_{1}^{3} \frac{( 3 x - y )}{5} d y

What is the integral from 0 to 1 of sin(x)(1-x) ?
The integral from 0 to 1 of sin(x)(1-x) is -sin(1)+1