I am finding the area bounded by parabola and line using definite integral. When its about line y = x and parabola $y^2 = x$. I know the line passes through origin. But when we have given equation of line like x = 4y - 2 and parabola $x^2 = 4y$ I got confused how to draw diagram.

I have no problem in finding points using given equations.

x = 4y - 2 and $x^2 = 4y$

We have when x = 2, y = 1.

And when x = -1, y = -1/4.

But problem comes in drawing sketch for this. Also when we find area why we subtract the integral of line from parabola or parabola from line.

Edit -

In this question I have doubts in drawing graph when the equation of line in the form of ax - by + c = 0.

Second on finding area why we don't add integrals of both?

Please ask me for any additional information.

  • $\begingroup$ $y^x=2$ does not define a parabola, probably there's a typo. However, are you able to find things like the axis and the vertex of a parabola? They should give you hints on how to draw it. $\endgroup$ – user402433 Dec 29 '16 at 13:14
  • $\begingroup$ With the hope of avoiding duplication of effort, could you please clarify how this differs from Find the area of the region bounded by parabola and line.? In that post as well, it seems you're struggling with how to set up the area of a plane region (bounded by two or more graphs) as an integral. Is that the underlying question? Thank you. :) $\endgroup$ – Andrew D. Hwang Dec 29 '16 at 13:43
  • $\begingroup$ @Andrew D. Hwang I can't ask many doubts in that one question so I asked different question here. Now I updated it with my doubts. $\endgroup$ – user404716 Dec 29 '16 at 13:47

Disclaimer: This answer invokes infinitesimals in a squishy, pragmatic way. The results are correct, but justifying their correctness requires work beyond the scope of this question.

The fundamental idiom of integral calculus (a name I'm trying to popularize) states:

To calculate a total quantity (area, length, volume, work...), split the quantity into infinitesimal increments that can be computed using algebra, then add them up (i.e., integrate) to find the total.

Suppose we wish to calculate the area of the plane region $R$ bounded below and above by the graphs $y = f_{1}(x)$ and $y = f_{2}(x)$, and bounded to the left and right by the vertical lines $x = a$, $x = b$. More succintly, $R$ is the set of $(x, y)$ with $$ f_{1}(x) \leq y \leq f_{2}(x),\qquad a \leq x \leq b. $$

To apply the fundamental idiom, we slice $R$ into "thin vertical strips": For each $x$ between $a$ and $b$, we have $f_{1}(x) \leq y \leq f_{2}(x)$. The portion of $R$ "at $x$" (or, if you prefer, "between $x$ and $x + dx$") is a rectangle of height $f_{2}(x) - f_{1}(x)$ and width $dx$, and so has area $$ dA = (f_{2}(x) - f_{1}(x))\, dx. $$ The total area of $R$ is the sum (i.e., integral) of these increments: $$ \text{area} = \int_{a}^{b} (f_{2}(x) - f_{1}(x))\, dx. $$

Area of a plane region sliced into vertical or horizontal strips

Entirely similar ideas work if our region $R$ can be sliced easily into horizontal strips, i.e., if $R$ can be conveniently defined by inequalities $$ g_{1}(y) \leq x \leq g_{2}(y),\qquad c \leq y \leq d. $$ In this event, $$ \text{area} = \int_{c}^{d} (g_{2}(y) - g_{1}(y))\, dy. $$

As for sketching in practice: Draw the specified curves separately, then determine (by inspection or calculation) where they intersect. If necessary, decide whether to calculate the area by slicing the region into vertical or horizontal strips, and express the region accordingly using graphs of functions. The points of intersection give the limits of integration.

In more complicated examples, it may be necessary to split the region into sub-regions, and to use the preceding paragraph separately on each piece. (Commonly, for example, one wants the region between two graphs that cross in the interior of the interval of integration.)

In your specific example, you have the line $x = 4y - 2$ (which can be sketched by finding two points on the line), and the parabola $x^{2} = 4y$, which opens to the right, is symmetric across the $x$-axis, and passes through the origin and the point $(2, 1)$.

Since each curve is given in the form of $x$ as a function of $y$, horizontal slices are easiest. Your sketch will tell you

  • Which curve is the left edge, $x = g_{1}(y)$, and which is the right edge, $x = g_{2}(y)$;

  • What the limits of integration, $c \leq y \leq d$, are.

  • $\begingroup$ I can't understand still why we subtract two areas to find bounded area. $\endgroup$ – user404716 Dec 29 '16 at 17:41
  • $\begingroup$ The two curves specify the endpoints of a thin rectangular slice; the relevant dimension of the slice is the difference: top minus bottom, or right minus left. Do the accompanying diagrams not help? $\endgroup$ – Andrew D. Hwang Dec 29 '16 at 18:00
  • $\begingroup$ No its not helpful. $\endgroup$ – user404716 Dec 30 '16 at 5:20
  • $\begingroup$ Is there a particular part of the argument you don't understand? In itemized form, the argument boils down to: 1. If a region is cut into thin strips, the total area of the region is equal to the sum of the areas of the strips. 2. Each strip is (approximately, but nearly enough so that the conclusion is correct) a rectangle, with one side very small. 3. The distance between two points on a number line is the larger minus the smaller. $\endgroup$ – Andrew D. Hwang Dec 30 '16 at 12:55
  • $\begingroup$ Actually the terms and language used by you is very difficult to understand. $\endgroup$ – user404716 Dec 30 '16 at 18:39

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