Advertisements
Advertisements
Question
Find the area of the triangle whose vertices are (-1, 1), (0, 5) and (3, 2), using integration.
Advertisements
Solution
Let A (-1,1), B (0,5) and C (3,2)
The equation of line AB is
y -1 = `(5 -1)/(0+ 1) ("x"+1)`
y = `4"x" + 5`
The equation of line BC is
y - 5 = `(2 -5)/(3 -0) ("x" -0)`
y = `-"x"+5`
The equation of line CA is
y - 2 = `(1 -2)/(-1 -3) ("x" -3)`
y = `("x")/(4) + (5)/(4)`
Required area = Area of ΔABC
The equation of line CA is
y - 2 = `(1 -2)/(-1 -3) ("x" -3)`
y = `("x")/(4) + (5)/(4)`
Required area = Area of ΔABC
= `int_-1^0 (4x + 5) dx + int_0^3 (- x + 5) dx - int_-1^3 ( x/4 + 5/4) dx`
= `[ 2x^2 + 5x ]_-1^0 + [ -x^2/2 + 5x ]_0^3 - [ x^2/8 + 5x/4 ]_-1^3`
= `3 + 21/2 - 39/8 - 9/8`
= `15/2` sq. units
APPEARS IN
RELATED QUESTIONS
Evaluate :`int_(pi/6)^(pi/3) dx/(1+sqrtcotx)`
Evaluate : `intsin(x-a)/sin(x+a)dx`
Find the integrals of the function:
sin 3x cos 4x
Find the integrals of the function:
sin 4x sin 8x
Find the integrals of the function:
`cos x/(1 + cos x)`
Find the integrals of the function:
`(cos 2x - cos 2 alpha)/(cos x - cos alpha)`
Find the integrals of the function:
`(cos x - sinx)/(1+sin 2x)`
Find the integrals of the function:
tan4x
`int (sin^2x - cos^2 x)/(sin^2 x cos^2 x) dx` is equal to ______.
Find `int (sin^2 x - cos^2x)/(sin x cos x) dx`
Evaluate: `int_0^π (x sin x)/(1 + cos^2x) dx`.
Evaluate `int_0^(3/2) |x sin pix|dx`
Find `int((3 sin x - 2) cos x)/(13 - cos^2 x- 7 sin x) dx`
Find `int_ sin ("x" - a)/(sin ("x" + a )) d"x"`
Find:
`int"dx"/sqrt(5-4"x" - 2"x"^2)`
Integrate the function `cos("x + a")/sin("x + b")` w.r.t. x.
Find: `int sec^2 x /sqrt(tan^2 x+4) dx.`
Find: `intsqrt(1 - sin 2x) dx, pi/4 < x < pi/2`
Find: `int sin^-1 (2x) dx.`
Find `int "dx"/(2sin^2x + 5cos^2x)`
`int "e"^x (cosx - sinx)"d"x` is equal to ______.
Evaluate the following:
`int tan^2x sec^4 x"d"x`
Evaluate the following:
`int sqrt(1 + sinx)"d"x`
`int (x + sinx)/(1 + cosx) "d"x` is equal to ______.
