Q.Evaluate the definite integral $\int_{0}^{1}\left(x\,e^x+\sin\frac{\pi x}{4}\right)dx$

The key idea is to split the integral into two simpler parts and evaluate each separately using standard antiderivatives.

Step 1: Split the integral

Step 2: First integral — integration by parts

Let $u = x$, $dv = e^x dx$. Then $du = dx$, $v = e^x$.

Evaluating from $0$ to $1$:

Step 3: Second integral — direct integration

Evaluating from $0$ to $1$: …

The key here is to break the problem into manageable pieces. A sum inside an integral can always be split into separate integrals — that's linearity, one of the most useful properties of definite integrals. Once we do that, each piece falls to a standard method.

1. Split the integral using linearity:

2. Evaluate $\int_{0}^{1} x e^x \, dx$ using integration by parts. The product $x \cdot e^x$ is a classic case: let $u = x$ and $dv = e^x dx$. Then $du = dx$ and $v = e^x$. Integration by parts gives:

So:

Now evaluate from $0$ to $1$:

3. Evaluate $\int_{0}^{1} \sin\frac{\pi x}{4} \, dx$ using a simple substitution. Let $u = \frac{\pi x}{4}$, so $du = \frac{\pi}{4} dx$, or $dx = \frac{4}{\pi} du$. When $x = 0$, $u = 0$; when $x = 1$, $u = \frac{\pi}{4}$. The integral becomes:

The antiderivative of $\sin u$ is $-\cos u$, so: …

Method: Split a sum, then choose the right tool for each piece

When the integrand is a sum of unrelated functions, use linearity to separate it, then match each piece to its own standard technique.

Steps

Step 1: Separate by linearity.

Step 2: For a product like $x\,e^x$, use integration by parts.

Pick $u=x$ (the algebraic factor, so its derivative simplifies) and $dv=e^x\,dx$:

$\int u\,dv = uv - \int v\,du.$ …

Common Mistakes

Mistake 1: Choosing $u=e^x$ instead of $u=x$ in the by-parts for $x e^x$.

Why it's wrong: you want the algebraic factor $x$ to differentiate away, so take $u=x$, $dv=e^x dx$; the other choice makes the integral harder. Correct approach: $\int xe^x\,dx=(x-1)e^x$, giving $1$ on $[0,1]$.

Mistake 2: Forgetting the $\tfrac{4}{\pi}$ factor when integrating $\sin\tfrac{\pi x}{4}$. …