find the area bounded; `y=x^2-2x-3` and `y=x+1`The graph begins in the upper left slopes down to cross over -1 (xaxis) crosses over the yaxis at -3 contiunes down to -4 and curves back up to cross...

find the area bounded; `y=x^2-2x-3` and `y=x+1`

The graph begins in the upper left slopes down to cross over -1 (xaxis) crosses over the yaxis at -3 contiunes down to -4 and curves back up to cross over at 3 on xaxis, the curve is below the 1. The stright line begins in the lower left crossses over -1(xaxis, and 1 y axis)

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lemjay | High School Teacher | (Level 3) Senior Educator

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Yes, your graph is correct as shown below. (Graph of `y=x^2-2x-3`  is red and `y=x+1` is green.)

Base on the graph, the two equations intersect at points (-1,0) and (4,5). 

Or we can solve it using substitution method. Substitute `y=x+1` to   `y=x^2 - 2x - 3` .

      `y = x^2 - 2x -3`

`x + 1 = x^2 - 2x - 3`

     ` 0 = x^2 - 3x - 4`

Factor right side.

    `0 = (x-4)(x+1)`

Set each factor to zero. Then, solve for x.

`x-4 = 0`       and       `x + 1 = 0`

     `x = 4 `                       `x = -1`

Substitute the values of x to y=x+1 .

`x= 4`     ===>  `y = 4+1 = 5`

`x= -1`   ===>  ` y = -1+1 =0`

Hence, the intersection points of the two equations are (-1,0) and (4,5).

Then, use the formula for area between two curves.

`A = int_a^b (y_U-y_L)dx`

Note that the limits of the integral are the x-coordinates of the intersection points. Also, `y_U` refers to graph of line  (`y=x+1` )  and  `y_L` refers to the parabola (`y=x^2-2x-3` ).

`A = int_-1^4 [(x+1) - (x^2-2x-3)] dx = int_-1^4(-x^2+3x+4)dx`

`A = -x^3/3 + (3x^2)/2 + 4x|_-1^4`

`A = [-4^3/3+(3*4^2)/2 + 4(4)]-[-(-1)^3/3 + (3(-1)^2)/2 + 4(-1)] `

`A = (-64/3 + 48/2 + 16) - (1/3+3/2-4) = (-128/6 + 144/6 + 96/6) - (2/6 + 9/6 - 24/6)`

`A = 112/6 - (-13/6) = 112/6 + 13/6 = 125/6 = 20 5/6`

Hence, area bounded by ` y=x^2-2x-3` and `y=x+1` is 20 5/6

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