Question

Consider the Assertion (A) and Reason (R) given below:
Assertion (A): \(\displaystyle\int_0^t \sin x\, dx = 1 - \cos t\)
Reason (R): \(\sin x\) is continuous in any closed interval \([0, t]\).

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true and R is not the correct explanation of A
• C. A is true but R is false
• D. A is false but R is true

Hint:

Evaluate the integral using the antiderivative of \(\sin x\) and check whether continuity is what allows the integral to be evaluated via the Fundamental Theorem of Calculus.

Answer 1

The Correct Option is A

Step 1: Define $F(t) = \int_0^t \sin x \, dx$. Since $\sin x$ is continuous on $[0,t]$ (this is exactly Reason R), the Fundamental Theorem of Calculus, Part 1, applies and guarantees $F$ is differentiable with $F'(t) = \sin t$.

Step 2: Solve this simple differential equation: $F'(t) = \sin t$ gives $F(t) = -\cos t + C$ for some constant $C$.

Step 3: Use the initial condition $F(0) = \int_0^0 \sin x\,dx = 0$ to find $C$: $0 = -\cos 0 + C = -1 + C$, so $C = 1$.

Step 4: Hence $F(t) = 1 - \cos t$, which is exactly the assertion. Since the continuity condition in R was the essential ingredient used to invoke the Fundamental Theorem of Calculus and derive this result, R correctly explains A.

\[\boxed{\text{Both A and R are true and R is the correct explanation of A}}\]