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The area enclosed between the curve y=si...

The area enclosed between the curve `y=sin^(2)x and y=cos^(2)x` in the interval `0le x le pi` is _____ sq. units.

A

2 sq unit

B

`(1)/(2)` sq unit

C

1 sq unit

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
`y=sin^(2)xand y=cso^(2)x`
Solving `sin^(2)x=cos^(2)x`
`therefore" "x=pi//4,3pi//4`
Graph of functions is as shown in the following figure.

From the figure, the required area
`=int_(pi//4)^(3pi//4)(sin^(2)x-cos^(2)x)dx`
`=-int_(pi//4)^(3pi//4)cos2xdx=1`
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The area enclosed by the curves y=sin x+cos x and y=|cos x-sin x| over the interval [0,(pi)/(2)]

Knowledge Check

  • The area between the curve y=-x^(2)+4x and y=x^(2)-6x+5 over the interval 0 le x le 1 is

    A
    `(52)/(3)+5sqrt(15)`
    B
    `-(52)/(3)+5sqrt(15)`
    C
    `-(52)/(3)+3sqrt(15)`
    D
    `(47)/(3)+5sqrt(15)`

  • The area enclosed by the curves y= sin x+ cos x and y= |cos x-sinx| over the interval (0, (pi)/(2)) is

    A
    `4(sqrt(2)-1)`
    B
    `2sqrt(2)(sqrt(2)-1)`
    C
    `2(sqrt(2)+1)`
    D
    `2sqrt(2)(sqrt(2)+1)`

  • The area between the curves y=x^(2) and y=x^(1//3), -1 le x le 1 is

    A
    `(1)/(2)`
    B
    2
    C
    `(3)/(4)`
    D
    `(3)/(2)`

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