Use calculus and find the area between the lines 3x + y = 4, 2x - y + 4 = 0 and the x-axis
- print Print
- list Cite
Expert Answers
calendarEducator since 2010
write12,554 answers
starTop subjects are Math, Science, and Business
The area between the lines 3x + y = 4, 2x - y + 4 = 0 and the x-axis has to be determined using calculus. This is quite easily done without the use of calculus but as the problem states that calculus has to be used here is the method.
The graph of the three lines is:
It is seen that the lines 3x + y = 4, 2x - y + 4 = 0 intersect at the point (0, 4). The lines intersect the x-axis at (-2, 0) and (4/3, 0)
Divide the area into two parts, one lying under the line 3x + y = 4 and the other under the line 2x - y + 4 = 0
The required area is:
`int_(-2)^0 2x + 4 dx` +` int_(0)^(4/3) 4 - 3x dx`
=> `x^2 + 4x` `|_(-2)^0` + `4x - (3x^2)/2 |_(0)^(4/3)`
=> `0^2 - (-2)^2 + 4*2 - 4*0 + 4(4/3 - 0) - (3/2)((4/3)^2 - 0)`
=> `20/3`
The required area is `20/3`
Related Questions
- Sketch the region enclosed by the given curves. y = 4 cos 2x, y = 4 − 4 cos 2x, 0 ≤ x ≤ π/2...
- 1 Educator Answer
- Find the vertices of triangle which sides are on the lines 2x+y-3=0; x-y-6=0; -2x+y-5=0?
- 2 Educator Answers
- cos x-sin 3x-cos 2x = 0
- 1 Educator Answer
- Find the area of the following region y=sin x and y=sin 2x from x=0 x=pi/2
- 1 Educator Answer
- x+ y = 5 2x-y = 4 solve x and y
- 2 Educator Answers