Area between a parabola and line? I learned that the distance between two curves is found but taking the integral of the upper curve subtracted by the integral of the lower curve, being evaluated at the intersecting points. I have equation: y = x − 1 y = x − 1 and y2 = 2x + 6 y 2 = 2 x + 6.

If −3 ≤ x ≤ 2 − 5–√ − 3 ≤ x ≤ 2 − 5 then both the upper and lower curves are parts of the parabola. If 2 − 5–√ ≤ x ≤ 2 + 5–√ 2 − 5 ≤ x ≤ 2 + 5 then the line is the lower curve and the upper one is y = 2x + 6− −−−−√ y = 2 x + 6.

We get a rightward-opening parabola and a line. The curves meet at (−1, −2) (− 1, − 2) and (5, 4) (5, 4). Now we have several options: Option 1: Integrate with respect to y y. Our "right" curve is x = y + 1 x = y + 1, and our "left" curve is x = y2−6 2 x = y 2 − 6 2.

The right formula is: 8 3|m|3 a(a − mc)3− −−−−−−−−√. area = ∫y2 y1 (y − c m − y2 4a) dy, my2 − 4ay + 4ac = 0. the area of a parabolic segment is 4/3 that of the inscribed triangle, having as vertex the projection, parallel to the axis, of the midpoint of the chord on the parabola.