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Application Of Integrals

Question
CBSEENMA12035753
Wired Faculty App

Using integration, find the area bounded by the curve x2 = 4y and the line x = 4y – 2.

Solution

The shaded area OBAO represents the area bounded by the curve x2 = 4y and the line x = 4y – 2.

Let A and B be the points of intersection of the line and parabola.
Co-ordinates of point A are open parentheses negative 1 comma space 1 fourth close parentheses. space Co-ordinates of point B are (2, 1).
Area OBAO = Area OBCO + Area OACO   ...(1)
Area OBCO = 
equals integral subscript 0 superscript 2 fraction numerator straight x plus 2 over denominator 4 end fraction dx minus integral subscript 0 superscript 2 straight x squared over 4 dx equals 1 fourth open square brackets straight x squared over 2 plus 2 straight x close square brackets subscript 0 superscript 2 minus 1 fourth open square brackets straight x cubed over 3 close square brackets subscript 0 superscript 2 equals 1 fourth open square brackets 2 plus 4 close square brackets minus 1 fourth open square brackets 8 over 3 close square brackets equals 3 over 2 minus 2 over 3 equals 5 over 6
Area OACO = 
  equals integral subscript negative 1 end subscript superscript 0 fraction numerator straight x plus 2 over denominator 4 end fraction dx minus integral subscript negative 1 end subscript superscript 0 straight x squared over 4 dx equals 1 fourth open square brackets straight x squared over 2 plus 2 straight x close square brackets subscript negative 1 end subscript superscript 0 space minus space 1 fourth open square brackets straight x cubed over 3 close square brackets subscript negative 1 end subscript superscript 0 equals 1 fourth open square brackets negative fraction numerator left parenthesis negative 1 right parenthesis squared over denominator 2 end fraction minus 2 left parenthesis negative 1 right parenthesis close square brackets minus 1 fourth open square brackets negative open parentheses negative 1 close parentheses cubed over 3 close square brackets equals 1 fourth open square brackets negative 1 half plus 2 close square brackets minus 1 fourth open square brackets 1 third close square brackets equals 3 over 8 minus 1 over 12 equals 7 over 24
Therefore, required area = open parentheses 5 over 6 plus 7 over 24 close parentheses space equals 9 over 8 sq. units

Some More Questions From Application of Integrals Chapter

A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is ` 100 and that on a bracelet is ` 300. Formulate on L.P.P. for finding how many of each should be produced daily to maximize the profit? It is being given that at least one of each must be produced.

Solve the following L.P.P. graphically :
Minimise Z = 5x + 10y
Subject to x + 2y ≤ 120
Constraints x + y ≥ 60
x – 2y ≥ 0
and x, y ≥ 0

The line L1: y = x = 0 and L2: 2x + y = 0 intersect the line L3: y + 2 = 0 at P and Q respectively. The bisectorof the acute angle between L1 and L2 intersects L3 at R.
Statement-1: The ratio PR: RQ equals 2√2:√5
Statement-2: In any triangle, the bisector of an angle divides the triangle into two similar triangles.

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