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`4:5``5:8``2:3``3:2`

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BTranscript

Time | Transcript |
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00:00 - 00:59 | Ravana Hue today's question is the area of the region bounded by the e to the power 4 y square is equals to 2 a minus x x to the power 5 is to that of the circle whose radius is given by the ratio so the given curve is a to the power 4 y square is equals to 2 a.m. - X into x to the power 5 and if it cut off the x-axis when the value of Y is equals to zero Janpad Y is equals to zero we get to a minus x x to the power 5 is equals to zero so the area bounded by the region a 1 is equals to integration 022 a.m. under root 2 A minus x x to the power 5 by 2 upon |

01:00 - 01:59 | a square B can we take admission we can write a test by square equals to 2 a.m. - 6 x to the power 5 EP 1842 the value of Y is become under root 2 minus x x to the power 5 by 2 upon a square so we can put x is equals to 2 sin square theta so the value of DX become 4A sin theta cos theta and sin theta so we can put it here we get and the limit with start from zero it remains zero but who is limit start from to A it becomes fire by to so the value of a 1 is equals to zero to pi by 2 and we can put the values route to A cos theta to a to the power 5 right to |

02:00 - 02:59 | sin 5 theta into 4 a sin theta cos theta theta because we can put the value of access to a sin square theta upon a square so so we can take the whole constant Ram out of the integral with his 32 because 222 to the power 5 by 2 becomes 32 and a square and the integration 0 to pi by 2 and overcome sin 5 theta and sin theta becomes sin to the power 6 theta and Cos theta and Cos theta become cos square theta theta so we can solve 32 a square 5 into 3 into 1 into |

03:00 - 03:59 | 1 upon 4 into 6 into 2 into 8 and 2 pi by 2 by the formula in which we can as sum the powers of the trigonometric function and take the autumn in the upper side and the event time and as the denominator so we can solve it so we get even is equals to 5 buy 85 a square and the area of the circle is which is a 2 is equals to a square and the radius is a source of squares of the ratio of even Upon A to is 5 by 8 5 a square upon 5 a square by a square and Square please cancel out said become 5 by 28 it is in the ratio |

04:00 - 04:59 | 5 by 8 thank you |

**Area as a definite integral**

**Algorithm for area calculation using horizontal strips**

**Area calculation using horizontal strips for modulus function**

**Impact of sign on area definition change according to sign of function**

**Find the area of the region bounded by the parabola `y^2 = 2x` and the line `x - y = 4`**

**Examples: Find the area of the region bounded by the curve `y^2 = 2y - x` and the y-axis.**

**Area of symmetrical graphs**

**Find the area of the region bounded by the ellipse `x^2 / a^2 + y^2 / b^2 = 1`**

**Integration of sin and cos function in different intervals**

**Examples: f(x) = x for x > 0 and f(x) = x^2 for x < 0; find area with x axis from x = -2 and x = 3.**