Q.Prove that $\int_{0}^{1}x\,e^x\,dx=1$

The key idea is integration by parts, which reverses the product rule for differentiation.

We set $u = x$ and $dv = e^x\,dx$. Then $du = dx$ and $v = e^x$.

Applying the formula $\int u\,dv = uv - \int v\,du$:

$\int_0^1 x e^x\,dx = \left[ x e^x \right]_0^1 - \int_0^1 e^x\,dx$ …

The core idea here is that we have a product of two functions: $x$ (a polynomial) and $e^x$ (an exponential). When you see a product like this, your first instinct should be integration by parts. Why? Because the derivative of $x$ is $1$, which is simpler, and the integral of $e^x$ is $e^x$, which is no harder. Integration by parts lets us trade a complicated product for a simpler one.

The formula for integration by parts is:

Think of it as the product rule for derivatives, but rearranged for integrals.

Let’s apply it step by step.

1. Choose $u$ and $dv$ wisely.

   We want $u$ to become simpler when differentiated, and $dv$ to be easy to integrate.

   Set $u = x$ and $dv = e^x \, dx$.

   Then $du = dx$ (the derivative of $x$ is $1$) and $v = e^x$ (the integral of $e^x$ is itself).

2. Plug into the formula.

3. Evaluate the boundary term.

   At $x = 1$: $1 \cdot e^1 = e$.

   At $x = 0$: $0 \cdot e^0 = 0$.

   So $[x e^x]_{0}^{1} = e - 0 = e$.

4. Evaluate the remaining integral. …

Method: Integration by parts for polynomial $\times$ exponential

Use this for $\int x^{n}e^{x}\,dx$ and similar products: choose the polynomial as $u$ so differentiating it lowers the degree.

Steps

Step 1: Choose $u$ and $dv$ by LIATE.

Take $u=x$ (algebraic, simplifies on differentiating) and $dv=e^x\,dx$, so $du=dx$ and $v=e^x$.

Step 2: Apply the formula.

$\int u\,dv=uv-\int v\,du=xe^x-\int e^x\,dx.$ …

Common Mistakes

Mistake 1: Wrong choice of $u$ and $dv$.

Why it's wrong: choosing $u=e^x$ makes the leftover integral harder; the polynomial $x$ should be $u$ so it simplifies on differentiating. Correct approach: $u=x$, $dv=e^x\,dx$.

Mistake 2: Sign error in the by-parts formula. …