Let $T\gt 0$ be a fixed number. If $f:\mathbb{R}\to\mathbb{R}$ is a continuous function such that \[ f(x+T)=f(x),\qquad x\in \mathbb{R}. \] If \[ I=\int_0^T f(x)\,dx, \] then \[ \int_0^{5T} f(2x)\,dx= \]

Question

If the solution of \[ \left( 1 + 2e^\frac{x}{y} \right) dx + 2e^\frac{x}{y} \left( 1 - \frac{x}{y} \right) dy = 0 \] is \[ x + \lambda y e^\frac{x}{y} = c \quad \text{(where $c$ is an arbitrary constant), then $ \lambda $ is:} \]

Answer 1

Step 1: State the given information.
$f$ is continuous with period $T > 0$: $f(x+T) = f(x)$ for all $x$. We have $I = \int_0^T f(x)\,dx$. We want $J = \int_0^{5T} f(2x)\,dx$.
Step 2: Substitute $u = 2x$ in $J$.
Let $u = 2x$, so $du = 2\,dx$, $dx = du/2$. When $x = 0$, $u = 0$; when $x = 5T$, $u = 10T$. So $J = \frac{1}{2}\int_0^{10T} f(u)\,du$.
Step 3: Use the periodicity property of definite integrals.
A key property states: if $f$ has period $T$, then for any positive integer $n$, $\int_0^{nT} f(u)\,du = n\int_0^T f(u)\,du$. This follows by splitting $[0, nT]$ into $n$ subintervals each of length $T$ and using periodicity.
Step 4: Apply the property with $n = 10$.
$\int_0^{10T} f(u)\,du = 10\int_0^T f(u)\,du = 10I$.
Step 5: Substitute back to find $J$.
$J = \frac{1}{2} \cdot 10I = 5I$.
Step 6: State the final answer.
\[\boxed{5I}\]

Let \(T\gt 0\) be a fixed number. If \(f:\mathbb{R}\to\mathbb{R}\) is a continuous function such that \[ f(x+T)=f(x),\qquad x\in \mathbb{R}. \] If \[ I=\int_0^T f(x)\,dx, \] then \[ \int_0^{5T} f(2x)\,dx= \]

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on Some Properties of Definite Integrals

Questions Asked in AP EAPCET exam