Integral Calculus
Bounds on ∫sinx/x dx
Grade Class 12
Question:
Value of $\displaystyle\int_0^1 \frac{\sin x}{x}\,dx$ lies in the interval
$\left(\dfrac{1695}{1800},\dfrac{1699}{1800}\right)$
$\left(\dfrac{1699}{1800},\dfrac{1703}{1800}\right)$
$\left(\dfrac{1703}{1800},\dfrac{1707}{1800}\right)$
$\left(\dfrac{1707}{1800},\dfrac{1711}{1800}\right)$
Step-by-Step Solution
Key Concept: Use Taylor expansion $\sin x/x=1-x^2/6+x^4/120-\ldots$ and integrate term by term: $\int_0^1(\sin x/x)dx=1-1/18+1/600-\ldots$ Bound between two partial sums.
By alternating series bounds: $\int_0^1\frac{\sin x}{x}dx\in(1699/1800,1703/1800)$.
Correct Answer: 2