Definite Integration
Integral Calculus-2
star_batch_jee_advanced_2025
Grade None
Question:
Let $T > 0$ be a fixed real number. Suppose $f(x)$ is a continuous function for all $x \in \mathbb{R}, f(x+T) = f(x)$. If $I = \int_0^T f(x)dx$ then:
\int_5^{5+5T} f(x)dx = 5I
\int_5^{5+5T} f(2x)dx = 10I
\int_5^{5+5T} f(3x)dx = 5I
\int_5^{5+5T} f(3x)dx = 15I
Step-by-Step Solution
Key Concept: For periodic functions, $\int_a^{a+nT} f(x)dx = nI$ regardless of the starting point $a$, and substitution $u = kx$ introduces a factor of $1/k$ that reduces the total integral.
Since $f(x)$ is periodic with period $T$, we have $\int_a^{a+nT} f(x)dx = nI$ for any real $a$ and positive integer $n$. For option 1: $\int_5^{5+5T} f(x)dx = 5\int_0^T f(x)dx = 5I$ ✓. For option 3: Let $u = 3x$, so $du = 3dx$. Then $\int_5^{5+5T} f(3x)dx = \frac{1}{3}\int_{15}^{15+15T} f(u)du = \frac{1}{3} \cdot 15I = 5I$ ✓. For options 2 and 4: The substitution $u = 2x$ gives $\int_5^{5+5T} f(2x)dx = \frac{1}{2}\int_{10}^{10+10T} f(u)du = 5I$ (not $10I$), and similarly option 4 gives $5I$ (not $15I$).
Correct Answer: 1,3