Question

Ifis continuous onand, show that

\(\int\limits_0^\infty {f'(x)dx = } - f(0)\)

Short answer

Hence proved the given expression\(\int\limits_0^\infty {f'(x)dx} = \mathop {\lim }\limits_{t \to \infty } \int\limits_0^t {f'(x)dx} \)

Step-by-step solution

Step 1: Solution of given expression

Use the definition of improper integrals to find its equivalent limit problem

\(\int\limits_0^\infty {f'(x)dx} = \mathop {\lim }\limits_{t \to \infty } \int\limits_0^t {f'(x)dx} \)

Step 2: Explanation of this improper integral by using fundamental theorem

\(\begin{aligned}{l}\mathop {\lim }\limits_{t \to \infty } \int\limits_0^t {f'(x)dx} = \mathop {\lim }\limits_{t \to \infty } (f(t) - f(0))\\ = \mathop {\lim }\limits_{t \to \infty } f(t) - \mathop {\lim }\limits_{t \to \infty } f(0)\\ = 0 - f(0)\\ = - f(0)\end{aligned}\)

Step 3: Final Proof

Consider the steps

The equation shows that successfully

\(\int\limits_0^\infty {f'(x)dx} = - f(0)\)