What is the integral from x to 1 of 2xsin(x) ?

The integral from x to 1 of 2xsin(x) is 2(-cos(1)+sin(1)+xcos(x)-sin(x))

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integral from x to 1 of 2xsin(x)

\int_{x}^{1} 2 x sin (x) d x

2 (- cos (1) + sin (1) + x cos (x) - sin (x))

Solution steps

\int_{x}^{1} 2 x sin (x) d x

Take the constant out:

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

= 2 \cdot \int_{x}^{1} x sin (x) d x

Apply Integration By Parts

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

= 2 [- x cos (x) - \int - cos (x) d x]_{x}^{1}

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

= 2 [- x cos (x) - (- sin (x))]_{x}^{1}

Simplify

= 2 [- x cos (x) + sin (x)]_{x}^{1}

Compute the boundaries:

- cos (1) + sin (1) + x cos (x) - sin (x)

= 2 (- cos (1) + sin (1) + x cos (x) - sin (x))

\int_{0}^{1} 4 r^{2} + 1 r d r

\int_{0}^{x} cos (3 t) d t

\int_{1}^{3} 5^{4 x} d x

\int_{1}^{2} 7 x d x

\int_{2}^{6} (2 x + 1) e^{0.4 x} d x

What is the integral from x to 1 of 2xsin(x) ?
The integral from x to 1 of 2xsin(x) is 2(-cos(1)+sin(1)+xcos(x)-sin(x))