Q.Evaluate $\int_{-1}^{3/2} \left| x \sin(\pi x) \right|\, dx$

Concept and Intuition

The absolute value $|x-3|$ measures distance from $3$. On the interval $[-2,2]$, every $x$ is less than $3$, so $x-3$ is negative. For a negative number, $|x-3| = -(x-3)$. Thus the fraction $\frac{|x-3|}{x-3}$ becomes $\frac{-(x-3)}{x-3} = -1$ for every $x$ in $[-2,2]$. The integral of a constant $-1$ over an interval is just that constant times the length of the interval. No splitting, no sign changes — it’s a flat line.

Step-by-step reasoning

1. Determine the sign of $x-3$ on $[-2,2]$

   The interval runs from $-2$ to $2$. Since $3$ is larger than any number in this interval, $x-3$ is always negative. For example, at $x=2$, $2-3=-1$; at $x=-2$, $-2-3=-5$.

2. Simplify the absolute value

   For any real number $a$, $|a| = -a$ when $a < 0$. Here $a = x-3 < 0$, so

3. Simplify the integrand Substitute into the fraction:

As long as $x \neq 3$ (and $3$ is not in $[-2,2]$), we can cancel $x-3$, giving

4. Integrate the constant …

$x^2+2x-3=(x-1)(x+3)$ is $\le0$ on $[0,1]$ and $\ge0$ on $[1,2]$. With antiderivative $F(x)=\frac{x^3}{3}+x^2-3x$:

$\int_0^1|\cdot|=-\big[F(1)-F(0)\big]=-\left(-\tfrac53-0\right)=\tfrac53,$ …

Concept & Intuition

At first glance, this looks like a trivial application of the Fundamental Theorem of Calculus (FTC):

If we let $f(x) = \tan^{-1}(1/x)$, then the integral would seem to be

That would suggest option (D). But this is wrong — and the pitfall is subtle but crucial.

The FTC requires $f$ to be differentiable and its derivative to be integrable on the closed interval $[-1,1]$. But $f(x) = \tan^{-1}(1/x)$ is not continuous at $x=0$ (it has a jump), so the derivative $f'(x)$ does not exist at $x=0$ in the usual sense, and the integral must be treated as an improper integral. The naive endpoint subtraction misses the jump.

We must split the integral at the discontinuity, handle each piece properly, and then combine.

Step-by-step solution

1. Identify the discontinuity

   The function $f(x) = \tan^{-1}(1/x)$ is undefined at $x=0$. As $x \to 0^+$, $1/x \to +\infty$, so $\tan^{-1}(1/x) \to \pi/2$. As $x \to 0^-$, $1/x \to -\infty$, so $\tan^{-1}(1/x) \to -\pi/2$. Hence there is a jump of size $\pi$ at $x=0$.

2. Split the integral at the singularity

   Write the integral as an improper integral:

On each subinterval, $f$ is continuous and differentiable, so the FTC applies separately.

3. Evaluate each piece using the FTC For the right piece:

For the left piece:

Step 1 — Spot the design of the question.

The limits are $2$ and $8$, and $a + b = 2 + 8 = 10$ — exactly the number appearing inside $\sqrt{10-x}$. That is never a coincidence; it signals the standard complementary-integrand family

whose value is always $\dfrac{b-a}{2}$.

Step 2 — The property being used.

For any integrable $f$,

This is the king property of definite integrals: reflecting the interval about its midpoint leaves the value unchanged.

Step 3 — Apply it.

With the integrand written as

replacing $x \to 10 - x$ swaps $g(x)$ and $g(10-x)$, so

Step 4 — Add the two forms of the same integral. …

The key insight is that powers of tangent can be rewritten using the identity $\tan^2 x = \sec^2 x - 1$. This lets us break a high power into a lower power times $\sec^2 x$ (which integrates nicely) minus an even lower power. The result is a clean recurrence that makes sums like $I_{10} + I_8$ collapse to a simple number.

1. Set up the reduction. For any $n \ge 2$, write

Replace $\tan^2 x$ with $\sec^2 x - 1$:

2. Integrate the first term. Notice that $\frac{d}{dx}(\tan x) = \sec^2 x$, so substitute $u = \tan x$, $du = \sec^2 x \, dx$. When $x=0$, $u=0$; when $x=\pi/4$, $u=1$. Thus

3. Write the recurrence. The second term is just $I_{n-2}$. So for $n \ge 2$,

4. Apply the recurrence to the sum we need. We want $I_{10} + I_8$. Using the formula with $n=10$: …