Q.Evaluate the definite integral: $\int_{-1}^1 (x+1) \ dx$

The key idea is that the definite integral of a sum can be split, and the integral of an odd function over a symmetric interval is zero.

Step 1: Split the integral:

Step 2: The function $x$ is odd, and the interval $[-1, 1]$ is symmetric about zero, so:

Step 3: The integral of the constant $1$ over $[-1, 1]$ is the length of the interval:

Step 4: Adding the results:

The key insight here is that the interval $[-1, 1]$ is symmetric about $0$. When you have a sum of functions, you can often break the integral into parts and use symmetry to simplify calculations. The function $x+1$ is not purely odd or even, but it is the sum of an odd function ($x$) and an even function ($1$). Over a symmetric interval $[-a, a]$, the integral of an odd function is zero, while the integral of an even function is twice the integral from $0$ to $a$. This saves you from having to compute the antiderivative directly, though that also works perfectly.

Let’s go through it step by step.

1. Separate the integral into two parts

This is valid because the integral of a sum is the sum of the integrals.

2. Handle the odd part: $\int_{-1}^1 x \, dx$ The function $f(x) = x$ is odd, meaning $f(-x) = -f(x)$. For any odd function integrated over a symmetric interval $[-a, a]$, the result is zero.

3. Handle the even part: $\int_{-1}^1 1 \, dx$ The constant function $g(x) = 1$ is even, since $g(-x) = g(x)$. For an even function over $[-a, a]$, the integral equals twice the integral from $0$ to $a$:

Now $\int_0^1 1 \, dx$ is just the length of the interval from $0$ to $1$, which is $1$. So:

4. Combine the results

If you prefer the direct antiderivative method, it’s just as straightforward:

Evaluating from $-1$ to $1$:

Same answer, confirming our symmetry approach.

Method: Evaluate a definite integral by the Fundamental Theorem of Calculus

For $\int_{a}^{b} f(x)\,dx$, find any antiderivative $F$, then compute $F(b)-F(a)$ — no constant of integration needed.

Steps

Step 1: Integrate term by term to get an antiderivative. For a linear integrand $x+1$, use the power rule: $F(x)=\dfrac{x^{2}}{2}+x$.

Step 2: Write the evaluation bracket.

Step 3: Substitute the upper limit, then the lower, and subtract.

$F(b)-F(a),$ being careful with signs when the lower limit is negative.

Step 4: Simplify to a number. The definite integral is a value, not a function, and needs no $+C$.

Common Mistakes

Mistake 1: Adding $+C$ to a definite integral.

Why it's wrong: the constant cancels in $F(b)-F(a)$. Correct approach: drop $C$ for definite integrals.

Mistake 2: Sign slip at a negative lower limit.

Why it's wrong: at $x=-1$, $\dfrac{(-1)^{2}}{2}+(-1)=\dfrac12-1=-\dfrac12$; subtracting this adds $\dfrac12$. Correct approach: compute $F(-1)$ fully, then subtract with its sign.

Mistake 3: Assuming "symmetry" makes it zero.

Why it's wrong: $x+1$ is not an odd function about $0$ (the $+1$ is even), so the integral is not $0$. Correct approach: split $\int_{-1}^{1}x\,dx=0$ but $\int_{-1}^{1}1\,dx=2$, giving $2$.