What is the integral from 2 to 4 of tcos(t) ?

The integral from 2 to 4 of tcos(t) is 4sin(4)+cos(4)-2sin(2)-cos(2)

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integral from 2 to 4 of tcos(t)

\int_{2}^{4} t cos (t) d t

4 sin (4) + cos (4) - 2 sin (2) - cos (2)

Decimal Notation

- 5.08330 \dots

Solution steps

\int_{2}^{4} t cos (t) d t

Apply Integration By Parts

: [t sin (t) - \int sin (t) d t]_{2}^{4}

= [t sin (t) - \int sin (t) d t]_{2}^{4}

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

= [t sin (t) - (- cos (t))]_{2}^{4}

Simplify

= [t sin (t) + cos (t)]_{2}^{4}

Compute the boundaries:

4 sin (4) + cos (4) - 2 sin (2) - cos (2)

= 4 sin (4) + cos (4) - 2 sin (2) - cos (2)

\frac{d}{d x} (- 2 (2 x^{2} cos (x^{2}) + sin (x^{2})))

x \to 1 lim (\frac{x ^{7} - x}{x - 1})

f (x) = 4 + x - 2 x^{2} - 3 x^{3}, at x = 2

x \to 1 lim (\frac{2}{x ^{3} - 1})

\frac{d}{d x} (sin (x^{c o s (x)}))

What is the integral from 2 to 4 of tcos(t) ?
The integral from 2 to 4 of tcos(t) is 4sin(4)+cos(4)-2sin(2)-cos(2)