Q.Choose the correct answer: If $f(x)=\int_{0}^{x}t\sin t\,dt$, then $f'(x)$ is (A) $\cos x+x\sin x$ (B) $x\sin x$ (C) $x\cos x$ (D) $\sin x+x\cos x$

The key idea is the First Fundamental Theorem of Calculus: if $F(x) = \int_a^x g(t)\,dt$, then $F'(x) = g(x)$.

Here, $f(x) = \int_0^x t \sin t \, dt$, so the integrand is $g(t) = t \sin t$. …

The core idea here is the First Fundamental Theorem of Calculus (FTC). It tells us that if you define a function as an integral from a constant to a variable, then the derivative of that function is just the integrand evaluated at that variable. In other words, differentiation undoes the integration from a fixed lower limit.

Why does this work? Imagine the integral as an area accumulator. As $x$ moves a tiny bit to the right, the area added is approximately the height of the curve at $x$ (which is $x \sin x$) times the tiny width. That height is exactly the rate at which the total area changes — i.e., the derivative.

Let’s apply this cleanly.

1. State the given function.

   We have $f(x) = \int_{0}^{x} t \sin t \, dt$. The lower limit is a constant (0), and the upper limit is the variable $x$.

2. Recall the First Fundamental Theorem of Calculus.

   If $F(x) = \int_{a}^{x} g(t) \, dt$, then $F'(x) = g(x)$, provided $g$ is continuous at $x$.

   Here, $g(t) = t \sin t$, which is continuous everywhere (product of continuous functions).

3. Apply the theorem directly. …

Method: Differentiating an integral with a variable limit — the First Fundamental Theorem

When a function is defined as an integral with a variable upper limit, its derivative is simply the integrand evaluated at that limit — no integration is needed.

Steps

Step 1: Identify the structure $F(x)=\displaystyle\int_a^{x} g(t)\,dt$.

The lower limit is a constant; the upper limit is the variable of differentiation.

Step 2: Apply the First Fundamental Theorem of Calculus.

Just replace the dummy variable $t$ by $x$ in the integrand. …

Common Mistakes

Mistake 1: Trying to compute the integral $\int_0^x t\sin t\,dt$ first, then differentiating.

Why it's wrong: the Fundamental Theorem lets you skip integration entirely — differentiating just returns the integrand at $x$. Correct approach: read off $f'(x)=x\sin x$ directly.

Mistake 2: Applying the product rule to $x\sin x$ as if differentiating it. …