If u(x) and v(x) are differentiable on [a, b], prove that

If $u(x)$ and $v(x)$ are differentiable on $[a, b]$, prove that $\int_a^b u(x) v'(x) \, dx = \left.u(x) v(x)\right|_a^b - \int_a^b v(x) u'(x) \, dx$ (Integration by Parts for Definite Integrals).

This is the formula for integration by parts for definite integrals. We will prove it using the product rule for derivatives.

Step 1: Start with the product rule

\[\frac{d}{dx}[u(x) v(x)] = u'(x) v(x) + u(x) v'(x)\]

Step 2: Rearrange to isolate $u(x) v'(x)$

\[u(x) v'(x) = \frac{d}{dx}[u(x) v(x)] - v(x) u'(x)\]

Step 3: Integrate both sides from $a$ to $b$

\[\int_a^b u(x) v'(x) \, dx = \int_a^b \frac{d}{dx}[u(x) v(x)] \, dx - \int_a^b v(x) u'(x) \, dx\]

Step 4: Apply the Fundamental Theorem of Calculus

\[\int_a^b \frac{d}{dx}[u(x) v(x)] \, dx = \left[u(x) v(x)\right]_a^b = u(b)v(b) - u(a)v(a) = \left.u(x) v(x)\right|_a^b\]

Step 5: Substitute back

\[\int_a^b u(x) v'(x) \, dx = \left.u(x) v(x)\right|_a^b - \int_a^b v(x) u'(x) \, dx\]

Therefore, we have proven the formula:

\[\boxed{\int_a^b u(x) v'(x) \, dx = \left.u(x) v(x)\right|_a^b - \int_a^b v(x) u'(x) \, dx}\]

Compact notation:

Letting $dv = v'(x) \, dx$ and $du = u'(x) \, dx$:

\[\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du\]

Key conditions:

• $u(x)$ and $v(x)$ are differentiable on $[a, b]$
• $u'(x)$ and $v'(x)$ are integrable on $[a, b]$
• All the integrals involved exist

Geometric interpretation:

The formula states that the area under the curve $u(x) v'(x)$ from $a$ to $b$ equals the change in the product $u(x) v(x)$ from $a$ to $b$, minus the area under the curve $v(x) u'(x)$.

Example application:

Verify with $\int_0^1 x e^x \, dx$: let $u=x$, $v'=e^x$, so $u'=1$, $v=e^x$:

\[\int_0^1 x e^x \, dx = \left[x e^x\right]_0^1 - \int_0^1 e^x \, dx = e - \left[e^x\right]_0^1 = e - (e-1) = 1\]

This demonstrates how integration by parts transforms a product into a simpler integral.

Historical note:

Integration by parts is one of the fundamental techniques of calculus, alongside substitution. It is particularly useful when the integrand is a product of two functions where one becomes simpler when differentiated and the other remains manageable when integrated.