\(\sqrt 5−2\sqrt 2+1\)
\(\sqrt 5+2\sqrt 2−4.5\)
Step 1: The intersection point of cos x - sin x = sin x is given by:
tan x = 1/2
So, let ψ = tan-1(1/2)
Hence, sin ψ = 1/√5 and cos ψ = 2/√5
Step 2: The graph of the equation is shown below:
Step 3: Now, the area between the curves can be calculated using the integral:
Area = ∫ψπ/2 (sin x - |cos x - sin x|) dx
Step 4: Break the integral into two parts:
= ∫ψπ/4 (sin x - (cos x - sin x)) dx + ∫π/4π/2 (sin x - (sin x - cos x)) dx
Step 5: Simplify the integrals:
= ∫ψπ/4 (2 sin x - cos x) dx + ∫π/4π/2 cos x dx
Step 6: Now, evaluate the integrals:
= [-2 cos x - sin x]ψπ/4 + [sin x]π/4π/2
Step 7: Substituting the limits:
= -√2 1/√2 + 2 cos ψ + sin ψ + (1 - 1/√2)
Step 8: Final evaluation:
= -√2 1/√2 + 2 (2/√5) + (1/√5) + (1 - 1/√2)
Conclusion: Therefore, the area is:
√5 - 2√2 + 1
The area of the region enclosed between the curve \( y = |x| \), x-axis, \( x = -2 \)} and \( x = 2 \) is:
Let the area of the region \( \{(x, y) : 2y \leq x^2 + 3, \, y + |x| \leq 3, \, y \geq |x - 1|\} \) be \( A \). Then \( 6A \) is equal to:
If the area of the region $$ \{(x, y): |4 - x^2| \leq y \leq x^2, y \geq 0\} $$ is $ \frac{80\sqrt{2}}{\alpha - \beta} $, $ \alpha, \beta \in \mathbb{N} $, then $ \alpha + \beta $ is equal to:
Read More: Area under the curve formula