Maybe I was a little bit hasty when I wrote my answer. It is important whether the required area is bounded only by [tex] y=x^2 [/tex] and its normal which encloses
angle of 135°with x-axis, or that area is bounded by [tex] y=x^2 [/tex], its normal and x-axis. You wrote that the required area is bounded only by [tex] y=x^2 [/tex] and it's normal so I would interpret that the following area is required to be determined:
This sketch that I wrote demonstrates what your task states. [tex] y=x^2 [/tex] is a basic form of parabola that has it's vertex in (0,0), and its normal that encloses 135° with x-axis intersects that parabola at 2 points. Since your task states that the area required to be determined is bounded only by [tex] y=x^2 [/tex] and its normal, it should be the area 'between' those two curves. [tex] k_{normal}=\tan 135° [/tex]. Since tangent and normal are perpendicular, you can find out coefficient of direction of tangent from condition of verticality: [tex] k_{tangent}=-\dfrac{1}{k_{normal}} [/tex]. You also know that: [tex] k_{tangent}=f'(x_1) [/tex], where [tex] f(x)=x^2 [/tex]. Now you can find 'x' coordinate for intersection point(point on the right where normal/tangent intersects parabola). To find 'y' coordinate, just plug 'x' that you've just got into [tex] y_1=x_1^2 [/tex]. Now you have the intersection point: [tex] T(x_1,y_1) [/tex] (the one on the right).
Equation of normal is: [tex] y-y_1=-\dfrac{1}{f'(x_1)}(x-x_1) [/tex]. Once you've got the equation of normal you can get the points where it intersects [tex] y=x^2 [/tex]. You should already have one intersection point by now [tex] T(x_1,y_1) [/tex] (it's the one on the right) , and now you should find the second point where normal intersects parabola(the one on the left on my sketch). To do so, you equalise equation of normal with the equation of parabola(you insert 'y' of normal into the equation of parabola): [tex] y_{normal}=x^2 [/tex]. Those two points of intersection should be the integration limits in this case. To find the area bounded by [tex] y=x^2 [/tex] and its normal(135° with x-axis), you can use formula: [tex] \int\limits_a^b \mathrm{(g(x)-f(x))} \mathrm{d}x [/tex]. f(x) is the equation of your graph ( [tex] y=x^2 [/tex] ), g(x) is the equation of normal, 'a' is 'x' value of the intersection point on the left side, and 'b' is the 'x' value of the intersection point on the right.
You can write [tex] \int\limits_a^b \mathrm{(g(x)-f(x))} \mathrm{d}x [/tex] as: [tex] \int\limits_a^b \mathrm{g(x)} \mathrm{d}x - \int\limits_a^b \mathrm{f(x)} \mathrm{d}x [/tex]. What we did here is subtract the whole area that is under the graph [tex] y=x^2 [/tex](between those integration limits) from the area that is under the normal(also between those integration limits). That way we got area between those two curves.
P.S. [tex] k_{normal} [/tex] and [tex] k_{tangent} [/tex] are slopes of those lines( coefficients of direction). For example: y=2x+3.
Slope of this line is 2. In this case 'k' are not points.
If normal creates the angle of 135°with the x-axis, it's the angle that I marked on my sketch.
You can use the same formula [tex] \int\limits_a^b \mathrm{(g(x)-f(x))} \mathrm{d}x [/tex] to find out area between any two curves. 'a' and 'b' are the 'x' values of points where they intersect, and b>a, g(x) is the curve that is higher on the y-axis, and f(x) is the curve that is positioned lower(g(x) is above f(x)).
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