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Popular >

integral from 0 to u of v+x

Solution

\int_{0}^{u} v + x d v

Solution

ux + \frac{u ^{2}}{2}

Solution steps

\int_{0}^{u} v + x d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{u} x d v + \int_{0}^{u} v d v

\int_{0}^{u} x d v = ux

\int_{0}^{u} v d v = \frac{u ^{2}}{2}

= ux + \frac{u ^{2}}{2}

Popular Examples

(1-t^2)y^'-t^2y=(1+t)*(sqrt(1-t^2))

(1 - t^{2}) y^{^{'}} - t^{2} y = (1 + t) \cdot (1 - t^{2})

tangent of 3x^2+y^2=1,(1/2 ,(-1)/2)tangent of

3 x^{2} + y^{2} = 1, (\frac{1}{2}, \frac{- 1}{2})

y^{''}-y^'+12y=0

y^{^{''}} - y^{^{'}} + 12 y = 0

derivative of ((e^x)/(1-e^{-x)})

\frac{d}{d x} (\frac{( e ^{x} )}{1 - e ^{- x}})

integral of-x/(x^2+y^2)

\int - \frac{x}{x ^{2} + y ^{2}} d x

Frequently Asked Questions (FAQ)

What is the integral from 0 to u of v+x ?
The integral from 0 to u of v+x is ux+(u^2)/2