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The area bounded by the curve y = sec x, the x-axis and the lines x = 0 and x = π/4 is
- a)log(√2 + 1)
- b)log(√2 - 1)
- c)1/2 log 2
- d)√2
Correct answer is option 'A'. Can you explain this answer?
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The area bounded by the curve y = sec x, the x-axis and the lines x = ...
Explanation:
Understanding the problem:
Given the curve y = sec x, the x-axis, and the lines x = 0 and x = π/4, we need to find the area bounded by them.
Identifying the bounds:
The curve y = sec x intersects the x-axis at x = π/2. So, the area is bounded by x = 0, x = π/4, x = π/2, and the curve y = sec x.
Integration:
To find the area, we can integrate the function sec x with respect to x over the limits 0 to π/4.
∫ sec x dx = ln|sec x + tan x| evaluated from 0 to π/4
Plugging in the values, we get:
ln|sec(π/4) + tan(π/4)| - ln|sec(0) + tan(0)|
= ln|√2 + 1| - ln|1 + 0|
= ln|√2 + 1|
Final Answer:
Therefore, the area bounded by the curve y = sec x, the x-axis, and the lines x = 0 and x = π/4 is log(√2 + 1), which corresponds to option 'A'.
Understanding the problem:
Given the curve y = sec x, the x-axis, and the lines x = 0 and x = π/4, we need to find the area bounded by them.
Identifying the bounds:
The curve y = sec x intersects the x-axis at x = π/2. So, the area is bounded by x = 0, x = π/4, x = π/2, and the curve y = sec x.
Integration:
To find the area, we can integrate the function sec x with respect to x over the limits 0 to π/4.
∫ sec x dx = ln|sec x + tan x| evaluated from 0 to π/4
Plugging in the values, we get:
ln|sec(π/4) + tan(π/4)| - ln|sec(0) + tan(0)|
= ln|√2 + 1| - ln|1 + 0|
= ln|√2 + 1|
Final Answer:
Therefore, the area bounded by the curve y = sec x, the x-axis, and the lines x = 0 and x = π/4 is log(√2 + 1), which corresponds to option 'A'.
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