Evaluate: int u frac{d^2v}{dx^2} dx - int v f | vedclass

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Question

(A) Using integration by parts for both integrals: $\int u \frac{d^2v}{dx^2} dx = u \frac{dv}{dx} - \int \frac{du}{dx} \frac{dv}{dx} dx$ $\int v \frac

Vedclass · Accepted Answer

$u \frac{dv}{dx} - v \frac{du}{dx} + C$

Vedclass · Answer

(A) Using integration by parts for both integrals: $\int u \frac{d^2v}{dx^2} dx = u \frac{dv}{dx} - \int \frac{du}{dx} \frac{dv}{dx} dx$ $\int v \frac{d^2u}{dx^2} dx = v \frac{du}{dx} - \int \frac{dv}{dx} \frac{du}{dx} dx$ Subtracting the two expressions: $\int u \frac{d^2v}{dx^2} dx - \int v \frac{d^2u}{dx^2} dx = (u \frac{dv}{dx} - \int \frac{du}{dx} \frac{dv}{dx} dx) - (v \frac{du}{dx} - \int \frac{dv}{dx} \frac{du}{dx} dx) + C$ $= u \frac{dv}{dx} - v \frac{du}{dx} - \int \frac{du}{dx} \frac{dv}{dx} dx + \int \frac{du}{dx} \frac{dv}{dx} dx + C$ $= u \frac{dv}{dx} - v \frac{du}{dx} + C$

Evaluate: $\int u \frac{d^2v}{dx^2} dx - \int v \frac{d^2u}{dx^2} dx$

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