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The area of the figure bounded by the curves y = cos x and y = sin x and the ordinates x = 0 and x = π/4 is
- a)√2 - 1
- b)√2 + 1
- c)1/√2 (√2 - 1
- d)1/√2
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
The area of the figure bounded by the curves y = cos x and y = sin x a...
The curves y = cos x and y = sin x intersect at x = π/4 and x = 5π/4. We can find the area of the bounded figure by finding the areas of the two regions separately and then adding them.
First, consider the region between the curves from x = 0 to x = π/4. We can set up an integral to find the area:
∫[0,π/4] (cos x - sin x) dx
Using integration by substitution with u = sin x, du = cos x dx, we can simplify this to:
∫[0,1] (1 - u) du = [u - u^2/2] from 0 to 1 = 1/2
So the area of this region is 1/2.
Next, consider the region between the curves from x = π/4 to x = 2π. We can set up a similar integral:
∫[π/4,2π] (cos x - sin x) dx
Using integration by substitution with u = cos x, du = -sin x dx, we can simplify this to:
-∫[1/√2,0] (1 - u) du = [u - u^2/2] from 1/√2 to 0 = 1/2 - 1/√2
So the area of this region is 1/2 - 1/√2.
Adding the areas of the two regions together, we get:
1/2 + 1/2 - 1/√2 = √2 - 1
So the area of the bounded figure is √2 - 1 square units.
First, consider the region between the curves from x = 0 to x = π/4. We can set up an integral to find the area:
∫[0,π/4] (cos x - sin x) dx
Using integration by substitution with u = sin x, du = cos x dx, we can simplify this to:
∫[0,1] (1 - u) du = [u - u^2/2] from 0 to 1 = 1/2
So the area of this region is 1/2.
Next, consider the region between the curves from x = π/4 to x = 2π. We can set up a similar integral:
∫[π/4,2π] (cos x - sin x) dx
Using integration by substitution with u = cos x, du = -sin x dx, we can simplify this to:
-∫[1/√2,0] (1 - u) du = [u - u^2/2] from 1/√2 to 0 = 1/2 - 1/√2
So the area of this region is 1/2 - 1/√2.
Adding the areas of the two regions together, we get:
1/2 + 1/2 - 1/√2 = √2 - 1
So the area of the bounded figure is √2 - 1 square units.
Community Answer
The area of the figure bounded by the curves y = cos x and y = sin x a...
$\frac{\pi}{4}$ can be found by integrating the difference between the two functions over that interval:
$\int_0^{\frac{\pi}{4}}(\cos x - \sin x) dx$
Using integration by substitution with $u = \sin x$, we can simplify this to:
$[-\cos x - \cos x]_0^{\frac{\pi}{4}} = -2\cos\frac{\pi}{4} + 2\cos 0 = -\sqrt{2} + 2$
So the area of the figure is $\boxed{\sqrt{2}-2}$.
$\int_0^{\frac{\pi}{4}}(\cos x - \sin x) dx$
Using integration by substitution with $u = \sin x$, we can simplify this to:
$[-\cos x - \cos x]_0^{\frac{\pi}{4}} = -2\cos\frac{\pi}{4} + 2\cos 0 = -\sqrt{2} + 2$
So the area of the figure is $\boxed{\sqrt{2}-2}$.
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The area of the figure bounded by the curves y = cos x and y = sin x and the ordinates x = 0 and x = π/4 isa)√2 - 1b)√2 + 1c)1/√2 (√2 - 1d)1/√2Correct answer is option 'A'. Can you explain this answer?
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