Fundamental theorems of integral calculus

At a Glance

• Frequency: 2 sub-parts across 2 of 13 years (2017, 2020)
• Priority tier: T3
• Marks (count): 15 (2)
• Average solve time: ~12 min
• Difficulty mix: medium 2
• Section: A | Dominant type: computation

Why This Chapter Matters

The two fundamental theorems of calculus are not just background — UPSC asks directly about their hypotheses. The 2017 question makes you diagnose exactly where FTC1 applies and where it fails (at the jump discontinuities of the floor function). The 2020 question uses the “King’s rule” reflection symmetry that collapses a hard integral into an easy one — a trick that appears repeatedly across years and papers. Both problems reward knowing the precise statement of a theorem over mechanical computation.

Minimum Theory

FTC1 (integral-function differentiability). If $f$ is Riemann integrable on $[a,b]$ and $F(x) = \int_a^x f(t)\,dt$, then $F$ is continuous on $[a,b]$. Moreover, at every point $x_0$ where $f$ is continuous, $F$ is differentiable and $F'(x_0) = f(x_0)$.

FTC2. If $F$ is differentiable on $[a,b]$ with $F' = f$ Riemann integrable on $[a,b]$, then $\int_a^b f(t)\,dt = F(b) - F(a)$.

Continuity of the integral function. If $f$ is bounded on $[a,b]$, then $F(x) = \int_a^x f(t)\,dt$ is Lipschitz (and hence continuous) on $[a,b]$ regardless of any discontinuities of $f$. Differentiability requires the integrand to be continuous at the point in question.

Symmetry (King’s rule). On $[0,a]$: $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$. This follows by the substitution $x \mapsto a-x$. Applying it and adding the two forms often collapses a complicated numerator into $1$ or another simple expression.

Question Archetypes

integral-function-smoothness (1 question(s); 2017)

Recognition Cues — The function is defined by $F(x) = \int_0^x g(t)\,dt$ where $g$ has known discontinuities (typically a piecewise or floor function). The question asks exactly where $F$ is differentiable and/or continuous but not differentiable. Apply FTC1: continuity is free (bounded integrand), differentiability requires continuity of $g$ at the point.

Solution Template

1. Identify discontinuities of $g$. Pinpoint every point where the integrand is discontinuous.
2. Prove $F$ is continuous everywhere. Use the Lipschitz bound: $|F(t)-F(t_0)| \le \|g\|_\infty |t-t_0| \to 0$.
3. Non-integer $t$: apply FTC1. At any $t$ where $g$ is continuous, $F'(t) = g(t)$.
4. Integer $t$: compute one-sided derivatives. In a left neighbourhood of $n$, the integrand has value $n-1$; in a right neighbourhood, value $n$. The one-sided derivatives are $F'_-(n) = n-1$ and $F'_+(n) = n$; they differ, so $F$ is not differentiable at $n$.
5. State answers clearly. Differentiable set; continuous-but-not-differentiable set.

Worked Example

2017 Paper 2, 2017-P2-Q2a (15 marks)

Let $f(t) = \displaystyle\int_0^t [x]\,dx$, where $[x]$ denotes the largest integer $\le x$. (i) Determine all real numbers $t$ at which $f$ is differentiable. (ii) Determine all real numbers $t$ at which $f$ is continuous but not differentiable.

Step 1 — $f$ is continuous everywhere.

The floor function $[x]$ is bounded on every bounded interval. For any $t_0$,

$|f(t)-f(t_0)| = \left|\int_{t_0}^t [x]\,dx\right| \le \sup_{[t_0-1,t_0+1]}|[x]| \cdot |t-t_0| \to 0.$

So $f$ is continuous on all of $\mathbb{R}$.

Step 2 — Differentiability at non-integer $t$.

If $t \notin \mathbb{Z}$, then $t$ lies in some open interval $(n, n+1)$ where $[x] = n$ is constant, hence continuous. By FTC1, $f'(t) = [t]$.

Step 3 — One-sided derivatives at integers.

At an integer $n$, compute:

$f'_-(n) = \lim_{h \to 0^-} \frac{f(n+h)-f(n)}{h} = [n-\epsilon] = n-1,$

$f'_+(n) = \lim_{h \to 0^+} \frac{f(n+h)-f(n)}{h} = [n+\epsilon] = n.$

Since $f'_+(n) - f'_-(n) = 1 \ne 0$, the derivative does not exist at any integer. This includes $n=0$: $f'_-(0) = -1 \ne 0 = f'_+(0)$.

Answers:

$\boxed{(i)\ f \text{ is differentiable at all } t \in \mathbb{R} \setminus \mathbb{Z},\text{ with } f'(t) = [t].}$

$\boxed{(ii)\ f \text{ is continuous but not differentiable at all } t \in \mathbb{Z}.}$

Common Traps

• $t = 0$ is not special. Students often state that $f$ is differentiable at $0$ because $f(0) = 0$. But the one-sided derivatives $f'_-(0) = -1$ and $f'_+(0) = 0$ differ, so $f$ fails to be differentiable there.
• Continuity does not fail. The integral of any bounded function is Lipschitz-continuous. Saying $f$ is discontinuous at integers is wrong; only differentiability fails.
• The slope jump is exactly 1 at every integer. The floor function increases by $1$ at each integer, giving a slope mismatch of $1$ regardless of which integer you check.

symmetry-integral (1 question(s); 2020)

Recognition Cues — A definite integral over $[0, \pi/2]$ where the integrand involves a sum $\sin x + \cos x$ in the denominator. The standard move is to apply the King substitution $x \mapsto \pi/2 - x$ and add the two forms. After the substitution, the numerator often simplifies via $\sin^2 x + \cos^2 x = 1$.

Solution Template

1. Name the integral $I$. Write $I = \int_0^{\pi/2} f(x)\,dx$.
2. Apply King’s rule. Set $J = \int_0^{\pi/2} f(\pi/2 - x)\,dx$; use the identities $\sin(\pi/2-x) = \cos x$, $\cos(\pi/2-x) = \sin x$ to simplify $J$.
3. Identify $I = J$ (if it holds). Check that the substitution just swaps $\sin \leftrightarrow \cos$ in the integrand.
4. Add $I + J$. The combined numerator usually becomes $\sin^2 x + \cos^2 x = 1$ (or another simple form), leaving a tractable integral.
5. Evaluate the resulting integral. Rewrite the denominator as $\sqrt{2}\sin(x + \pi/4)$ and integrate using $\int \csc t\,dt = \log|\tan(t/2)|$.
6. Simplify. Use $\tan(\pi/8) = \sqrt{2}-1$ and $\tan(3\pi/8) = \sqrt{2}+1$ to obtain the exact value.

Worked Example

2020 Paper 2, 2020-P2-Q4b (15 marks)

Show that $\displaystyle\int_0^{\pi/2}\frac{\sin^2 x}{\sin x+\cos x}\,dx=\frac{1}{\sqrt2}\log_e(1+\sqrt2)$.

Let $I = \int_0^{\pi/2}\dfrac{\sin^2 x}{\sin x+\cos x}\,dx$.

Step 1 — King’s rule. Substitute $x \mapsto \pi/2 - x$:

$J = \int_0^{\pi/2}\frac{\cos^2 x}{\cos x+\sin x}\,dx.$

Since the denominator is unchanged and the numerator swaps $\sin^2 \leftrightarrow \cos^2$, we get $I = J$.

Step 2 — Add.

$2I = I + J = \int_0^{\pi/2}\frac{\sin^2 x + \cos^2 x}{\sin x + \cos x}\,dx = \int_0^{\pi/2}\frac{dx}{\sin x + \cos x}.$

Step 3 — Evaluate $K = \int_0^{\pi/2} dx/(\sin x + \cos x)$.

Write $\sin x + \cos x = \sqrt{2}\sin(x+\pi/4)$, so

$K = \frac{1}{\sqrt{2}}\int_0^{\pi/2} \csc\!\left(x+\frac{\pi}{4}\right)dx.$

Substitute $t = x + \pi/4$, limits $\pi/4$ to $3\pi/4$:

$K = \frac{1}{\sqrt{2}}\Big[\log\big|\tan\tfrac{t}{2}\big|\Big]_{\pi/4}^{3\pi/4} = \frac{1}{\sqrt{2}}\log\frac{\tan(3\pi/8)}{\tan(\pi/8)}.$

Step 4 — Simplify. Using $\tan(\pi/8) = \sqrt{2}-1$ and $\tan(3\pi/8) = \sqrt{2}+1$:

$\frac{\tan(3\pi/8)}{\tan(\pi/8)} = \frac{\sqrt{2}+1}{\sqrt{2}-1} = (\sqrt{2}+1)^2.$

Therefore $K = \dfrac{1}{\sqrt{2}}\log(\sqrt{2}+1)^2 = \dfrac{2}{\sqrt{2}}\log(1+\sqrt{2}) = \sqrt{2}\log(1+\sqrt{2})$.

Step 5 — Finish.

$I = \frac{K}{2} = \frac{\sqrt{2}}{2}\log(1+\sqrt{2}) = \frac{1}{\sqrt{2}}\log(1+\sqrt{2}).$

$\boxed{\int_0^{\pi/2}\frac{\sin^2 x}{\sin x+\cos x}\,dx = \frac{1}{\sqrt{2}}\log_e(1+\sqrt{2})}$

Common Traps

• Skipping the King step. Trying to evaluate $\sin^2 x/(\sin x + \cos x)$ by Weierstrass substitution or partial fractions is extremely messy. The reflection $x \mapsto \pi/2 - x$ is the only clean entry point.
• Forgetting $\tan(\pi/8) = \sqrt{2}-1$. The final simplification depends on knowing or quickly deriving this half-angle value. Derive it with the half-angle formula: $\tan(\pi/8) = 1/(1+\sqrt{2}/1) = \sqrt{2}-1$ (rationalise $\tan(\pi/4)/(1+\sec(\pi/4))$). Under exam pressure, verify numerically: $\tan(22.5°) \approx 0.414 \approx \sqrt{2}-1$. ✓
• Arithmetic on $\frac{1}{2} \cdot \sqrt{2}$. Note $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$; these are the same. Write whichever form matches the target expression.

Marks-Aware Writing

A 15-mark answer for the FTC question requires: (1) an explicit Lipschitz/uniform continuity argument to establish global continuity, (2) FTC1 cited for non-integers with $f'(t) = [t]$ written clearly, (3) both one-sided limits computed at a general integer $n$ with the difference $= 1$ shown. Omitting the continuity proof or treating $t=0$ carelessly each costs marks.

For the symmetry integral: (1) define $I$ and $J$ and show $I = J$ by King’s rule, (2) add and simplify to get $2I = \int 1/(\sin x + \cos x)$, (3) evaluate via $\sqrt{2}\sin(x+\pi/4)$ and the $\csc$ antiderivative, (4) compute $\tan(\pi/8)$ and $\tan(3\pi/8)$ and carry through the logarithm identity. All four steps must appear for full marks.

Practice Set

• 2013-P2-Q1c (10 m) — — Hint: likely another FTC / continuity-of-integral question; establish where the integrand is continuous first.
• 2014-P2-Q2b (15 m) — — Hint: may involve a symmetry substitution or integral-function differentiability; identify the technique from the structure of the integrand.