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The area of the figure bounded by the curves y = ex, y = e-x & the straight line x = 1 is
- a)e + 1/e
- b)e - 1/e
- c)e + 1/e - 2
- d)None Of these
Correct answer is option 'C'. Can you explain this answer?
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The area of the figure bounded by the curves y = ex, y = e-x & the...
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The area of the figure bounded by the curves y = ex, y = e-x & the...
To find the area bounded by the curves y = ex and y = e-x, we need to find the points where they intersect.
Setting ex = e-x, we can take the natural logarithm of both sides to get ln(ex) = ln(e-x).
This simplifies to x = -x.
Adding x to both sides, we get 2x = 0, and dividing both sides by 2 gives x = 0.
So the curves intersect at the point (0, 1).
To find the area, we integrate the difference between the two curves from x = 0 to x = 1.
The area is given by the integral from 0 to 1 of (ex - e-x) dx.
This integral can be evaluated using the substitution method.
Let u = ex, then du = ex dx.
The integral becomes ∫(u - 1/u) du.
Integrating, we get (u^2/2 - ln|u|) evaluated from 0 to 1.
Plugging in the limits, we get ((1^2/2 - ln|1|) - (0^2/2 - ln|0|)).
This simplifies to (1/2 - 0) - (0 - undefined).
Since the natural logarithm of 0 is undefined, the area is 1/2.
Therefore, the area of the figure bounded by the curves y = ex and y = e-x is 1/2.
Setting ex = e-x, we can take the natural logarithm of both sides to get ln(ex) = ln(e-x).
This simplifies to x = -x.
Adding x to both sides, we get 2x = 0, and dividing both sides by 2 gives x = 0.
So the curves intersect at the point (0, 1).
To find the area, we integrate the difference between the two curves from x = 0 to x = 1.
The area is given by the integral from 0 to 1 of (ex - e-x) dx.
This integral can be evaluated using the substitution method.
Let u = ex, then du = ex dx.
The integral becomes ∫(u - 1/u) du.
Integrating, we get (u^2/2 - ln|u|) evaluated from 0 to 1.
Plugging in the limits, we get ((1^2/2 - ln|1|) - (0^2/2 - ln|0|)).
This simplifies to (1/2 - 0) - (0 - undefined).
Since the natural logarithm of 0 is undefined, the area is 1/2.
Therefore, the area of the figure bounded by the curves y = ex and y = e-x is 1/2.
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The area of the figure bounded by the curves y = ex, y = e-x & the straight line x = 1 isa)e +1/eb)e -1/ec)e +1/e- 2d)None Of theseCorrect answer is option 'C'. Can you explain this answer?
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