What is the area of the region bounded by the curve y = \sin x and the straight line y = \… — JAMB UTME

What is the area of the region bounded by the curve $y = \sin x$ and the straight line $y = \frac{2}{\pi}x$ in the first quadrant for $0 \leq x \leq \frac{\pi}{2}$ ?

$1 - \frac{\pi}{4}$ square unitsAnswer
B
$1 + \frac{\pi}{4}$ square units
C
$1 - \frac{\pi}{2}$ square units
D
$\frac{\pi}{4}$ square units$

Answer

The area of the enclosed region is

1 - \frac{\pi}{4}

square units.

The bounded area is calculated by taking the definite integral of the upper curve minus the lower curve over the interval

[0, \pi/2]

. The upper boundary is

y = \sin x

and the lower boundary is

y = \frac{2}{\pi}x

. Computing

\int_{0}^{\pi/2} (\sin x - \frac{2}{\pi}x) dx

gives

[-\cos x - \frac{x^2}{\pi}]_{0}^{\pi/2} = (0 - \frac{\pi}{4}) - (-1 - 0) = 1 - \frac{\pi}{4}

square units.

Step-by-Step Solution

Set up the definite integral for the area between the two curves

A = \int_{0}^{\pi/2} \left(\sin x - \frac{2}{\pi}x\right) dx

In the interval

[0, \pi/2]

, the curve

y = \sin x

lies above the straight line

y = \frac{2}{\pi}x

Find the antiderivative of each term

\int \sin x \, dx = -\cos x \quad \text{and} \quad \int \frac{2}{\pi}x \, dx = \frac{x^2}{\pi}

Integration rules for basic trigonometric and power functions.

Evaluate the definite integral from lower limit

x = 0

to upper limit

x = \frac{\pi}{2}

A = \left[-\cos x - \frac{x^2}{\pi}\right]_{0}^{\pi/2} = \left(-\cos\frac{\pi}{2} - \frac{(\pi/2)^2}{\pi}\right) - \left(-\cos 0 - \frac{0^2}{\pi}\right)

Apply the Fundamental Theorem of Calculus:

F(b) - F(a)

Simplify the numeric expression

A = \left(0 - \frac{\pi}{4}\right) - (-1 - 0) = -\frac{\pi}{4} + 1 = 1 - \frac{\pi}{4}

Since

\cos(\pi/2) = 0

and

\cos(0) = 1

Key Concept

Area bounded between curves using definite integration

Estimated Time:2m 0s

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