5.18 Proof of integration by parts formula

(Not examinable.) By the product rule, we have

$$(fg)^{\prime}(x)=f^{\prime}(x)g(x)+f(x)g^{\prime}(x),$$

so by integration, we have

$$\int_{a}^{b}(fg)^{\prime}(x)\,dx=\int_{a}^{b}f^{\prime}(x)g(x)\,dx+\int_{a}^{b}f(x)g^{\prime}(x)\,dx;$$

so by the fundamental theorem of calculus, we deduce

$$\bigl[f(x)g(x)\bigr]_{a}^{b}=\int_{a}^{b}f^{\prime}(x)g(x)\,dx+\int_{a}^{b}f(x)g^{\prime}(x)\,dx,$$

and by rearranging, we conclude

$$\int_{a}^{b}f(x)g^{\prime}(x)\,dx=\bigl[f(x)g(x)\bigr]_{a}^{b}-\int_{a}^{b}f^{\prime}(x)g(x)\,dx.$$