int (f(x) g^{prime prime}(x) - f^{prime prime | vedclass

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(C) We use the method of integration by parts for the given integral: $\int (f(x) g^{\prime \prime}(x) - f^{\prime \prime}(x) g(x)) \, dx$. First,cons

Vedclass · Accepted Answer

$f(x) g^{\prime}(x) - f^{\prime}(x) g(x) + C$

Vedclass · Answer

(C) We use the method of integration by parts for the given integral: $\int (f(x) g^{\prime \prime}(x) - f^{\prime \prime}(x) g(x)) \, dx$. First,consider the integral $\int f(x) g^{\prime \prime}(x) \, dx$. Using integration by parts,let $u = f(x)$ and $dv = g^{\prime \prime}(x) \, dx$. Then $du = f^{\prime}(x) \, dx$ and $v = g^{\prime}(x)$. So,$\int f(x) g^{\prime \prime}(x) \, dx = f(x) g^{\prime}(x) - \int f^{\prime}(x) g^{\prime}(x) \, dx$. Next,consider the integral $\int f^{\prime \prime}(x) g(x) \, dx$. Using integration by parts,let $u = g(x)$ and $dv = f^{\prime \prime}(x) \, dx$. Then $du = g^{\prime}(x) \, dx$ and $v = f^{\prime}(x)$. So,$\int f^{\prime \prime}(x) g(x) \, dx = g(x) f^{\prime}(x) - \int g^{\prime}(x) f^{\prime}(x) \, dx$. Subtracting the two results: $\int f(x) g^{\prime \prime}(x) \, dx - \int f^{\prime \prime}(x) g(x) \, dx = [f(x) g^{\prime}(x) - \int f^{\prime}(x) g^{\prime}(x) \, dx] - [g(x) f^{\prime}(x) - \int g^{\prime}(x) f^{\prime}(x) \, dx]$. The integral terms $\int f^{\prime}(x) g^{\prime}(x) \, dx$ cancel out. Thus,the result is $f(x) g^{\prime}(x) - f^{\prime}(x) g(x) + C$.

$\int (f(x) g^{\prime \prime}(x) - f^{\prime \prime}(x) g(x)) \, dx$ is equal to

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