Let's plot the graph:

x = var('x')
g = Graphics()
g += plot(x^3 + x^2, 0, 3, legend_label = "y = x^3 + x^2")
g += plot(x^3 - 1, 0, 3, legend_label = "x^3 - 1", color="green")
g.show()

Let's find the area between the curves:

\(A = \int_0^3 (x^3 + x^2) - (x^3 - 1)\)

from sage.symbolic.integration.integral import definite_integral
x = var('x')
definite_integral(x^2 + 1, x, 0, 3)

12

Let's plot the graph:

x = var('x')
g = Graphics()
g += plot(1/(x^2),1, 2, legend_label="$\dfrac{1}{x^2}$")
g += plot(1/(x^3), 1, 2, legend_label="$\dfrac{1}{x^3}$", color = "green")
g.show()

Let's find the area between the curves:

\(A = \int_1^2 \dfrac{1}{x^2} - \dfrac{1}{x^3}\)

x = var('x')
definite_integral(1/(x^2) - 1/(x^3), x, 1, 2)

1/8

Since no interval is given, let's compute the intersection point of the curves.

x = var('x')
solve([x^2 == x^3], x)

[x == 0, x == 1]

g1 = x^2
g2 = x^3

[g1(0), g1(1), g2(0), g2(1)]

[0, 1, 0, 1]

So the intersection points of the equations are \((0,0)\) and \((1,1)\).

Let's plot the graph between the intersection points:

x = var('x')
g = Graphics()
g += plot(x^2, 0, 1, legend_label = "x^2")
g += plot(x^3, 0, 1, legend_label = "x^3", color="green")
g.show()

Let's find the area between the curves:

\(A = \int_0^1 x^2 - x^3\)

x = var('x')
definite_integral((x^2) - (x^3), x, 0, 1)

1/12

Let's plot the graph:

x = var('x')
g = Graphics()
g += plot(1/2, pi/6, (5*pi/6), legend_label = "$\dfrac{1}{2}$")
g += plot(sin(x), pi/6, (5*pi/6), legend_label = "$\sin(x)$", color="green")
g.show()

Let's find the area between the curves:

\(A = \int_{\dfrac{\pi}{6}}^{\dfrac{5\pi}{6}} \sin x - (-\dfrac{1}{2})\)

definite_integral(sin(x) + (1/2), x, pi/6, 5*pi/6)

1/3*pi + sqrt(3)

Let's plot the graph:

x = var('x')
g = Graphics()
g += plot(sin(x), pi/4, (5*pi/4), legend_label = "$\sin(x)$")
g += plot(cos(x), pi/4, (5*pi/4), legend_label = "$\cos(x)$", color="green")
g.show()

Let's find the area between the curves:

\(A = \int_(\pi/4)^(5\pi/4) \sin x - \cos x dx\)

definite_integral(sin(x) - cos(x), x, pi/4, 5*pi/4)

2*sqrt(2)

The equation of y axis is \(x = 0\) Let's find the intersection points with the curve \(x = y^2 - 1\)

x,y = var('x y')
solve([y^2 - 1 == 0], y)

[y == -1, y == 1]

Let's plot the graph:

x,y = var('x y')
g = Graphics()
g += line([(0,0), (0,1), (0,-1)], legend_label = "x = 0")
g += plot(sqrt(x + 1), -1, 2, legend_label = "$\sqrt{x+1}$", color="green")
g += plot(-sqrt(x + 1), -1, 2, legend_label = "$-\sqrt{x+1}$", color="green")
g.show()

Let's find the area between the curves:

\(A = \int_{-1}^1 \sqrt{x+1} - (-\sqrt{x+1}) dx\)

\(= \int_{-1}^1 2\sqrt{x+1} dx\)

definite_integral(2*sqrt(x+1), x, -1, 1)

8/3*sqrt(2)

Since no interval is given, let's compute the intersection point of the curves.

x = var('x')
solve(sqrt(x) == x/2, x, algorithm='sympy')

[x == 0, x == 4]

Let's compute the points:

g1 = x/2
[g1(0), g1(4)]

[0, 2]

So the intersection points are \((0,0)\) and \((4,2)\)

Let's plot the graph:

  x,y = var('x y')
  g = Graphics()
  g += plot(sqrt(x), 0, 4, legend_label = "$\sqrt{x}$")
  g += plot(x/2, 0, 4, legend_label = "$\dfrac{x}{2}$", color="green")
  g.show()

Let's find the area between the curves:

\(A = \int_0^4 \sqrt{x} - (\dfrac{x}{2}) dx\)

\(= \int_0^4 \sqrt{x} - \dfrac{x}{2}\)

definite_integral(sqrt(x) - (x/2), x, 0, 4)

4/3

Since no interval is given, let's compute the intersection point of the curves.

x = var('x')
solve(x^3 - x == 3*x, x)

[x == -2, x == 2, x == 0]

Let's plot the graph:

  x,y = var('x y')
  g = Graphics()
  g += plot(x^3 - x, -2, 2, legend_label = "$x^3 - x$")
  g += plot(3*x, -2, 2, legend_label = "$3x$", color="green")
  g.show()

Let's find the area between the curves:

\(A = \int_{-2}^0 x^3 - x - 3x dx + \int_0^2 3x - (x^3 - x) dx\)

\(= \int_{-2}^0 x^3 - 4x dx + \int_0^2 4x - x^3 dx\)

definite_integral(x^3 - 4*x, x, -2, 0) + definite_integral(4*x - x^3, x, 0, 2)

8

I'm switching the \(x\) and \(y\) in the curves to simpify the problem.

x = var('x')
solve(x^2 + x == 3 - x, x)

[x == -3, x == 1]

Let's plot the graph:

  x = var('x')
  g = Graphics()
  g += plot(x^2 + x, -3, 1, legend_label = "$x^2 + x$")
  g += plot(3- x, -3, 1, legend_label = "$3-x$", color="green")
  g.show()

Let's find the area between the curves:

\(A = \int_{-3}^1 3 - x - (x^2 + x) dx\)

\(= \int_{-3}^1 3 - 2x - x^2 dx\)

definite_integral(3 - 2*x - x^2, x, -3, 1)

32/3

Since no interval is given, let's compute the intersection point of the curves.

x = var('x')
solve(x == 2*x, x)

[x == 0]

Intersection between \(y=x\) and \(y = 2x\) is at \(x = 0\)

x = var('x')
solve(x == 6 - x, x)

[x == 3]

Intersection between \(y=x\) and \(y = 6-x\) is at \(x = 3\)

x = var('x')
solve(2*x == 6 - x, x)

[x == 2]

Intersection between \(y=2x\) and \(y = 6-x\) is at \(x = 2\)

Let's plot the graph:

  x = var('x')
  g = Graphics()
  g += plot(x, 0, 3, legend_label = "$x$")
  g += plot(2*x, 0, 3, legend_label = "$2x$", color="red")
  g += plot(6 - x, 0, 3, legend_label = "$6-x$", color="green")
  g.show()

Let's find the area between the curves:

\(\int_0^2 2x - x dx + \int_2^3 6 - x - x dx\)

\(= \int_0^2 x dx + \int_2^3 6 - 2xdx\)

definite_integral(x, x, 0, 2) + definite_integral(6 - 2*x, x, 2, 3)

3

Since no interval is given, let's compute the intersection point of the curves.

solve(x^2 == sqrt(x), x, algorithm='sympy')

[[x == 0, x == 1],
 ConditionSet(x, Eq(-sqrt(x) + x**2, 0), FiniteSet(-1/2 - sqrt(3)*I/2, -1/2 + sqrt(3)*I/2))]

Let's plot the graph:

  x = var('x')
  g = Graphics()
  g += plot(x^2, 0, 1, legend_label = "$x^2$")
  g += plot(sqrt(x), 0, 1, legend_label = "$\sqrt{x}$", color="green")
  g.show()

Let's find the area between the curves:

\(A = \int_0^1 \sqrt{x} - x^2 dx\)

from sage.symbolic.integration.integral import definite_integral
definite_integral(sqrt(x) - (x^2), x, 0, 1)

1/3

Since no interval is given, let's compute the intersection point of the curves.

x = var('x')
solve(x == 2*(3*x - 2*(x^2)) - (3*x - (2*x^2))^2, x)

[x == (-1/2*I + 1), x == (1/2*I + 1), x == 1, x == 0]

Before plotting the graph, we need to understand the curve \(x = 2y - y^2\).

x,y = var('x y')
solve(x == 2*y - y^2, y)

[y == -sqrt(-x + 1) + 1, y == sqrt(-x + 1) + 1]

Let's plot the graph:

  x = var('x')
  g = Graphics()
  g += plot(3*x - 2*(x^2), 0, 1, legend_label = "$3x - 2x^2$")
  g += plot(sqrt(-x + 1) + 1, 0, 1, legend_label = "$\sqrt{-x + 1} + 1$", color="green")
  g += plot(-sqrt(-x + 1) + 1, 0, 1, legend_label = "$-\sqrt{-x + 1} + 1$", color="red")
  g.show()

Let's find the area between the curves:

\(A = \int_0^1 3x-2x^2 - (-\sqrt{-x + 1} + 1) dx\)

\(= \int_0^1 3x-2x^2 + \sqrt{-x + 1} - 1 dx\)

definite_integral(3*x - (2*x^2) + sqrt(-x + 1) - 1, x, 0, 1)

1/2

We have to find three pieces of area.

For the first piece of area:

\(= \int_0^1 \sqrt{x} - (-\sqrt{x}) dx\)

Computing it:

a = definite_integral(2*sqrt(x), x, 0, 1)
a

4/3

For the second piece of area:

\(= \int_1^2 \sqrt{x} - (x-2) dx\)

Computing it:

b = definite_integral(sqrt(x) - x + 2, x, 1, 2)
b

4/3*sqrt(2) - 1/6

For the third piece of area:

\(= \int_2^4 \sqrt{x} - (x-2) dx\)

Computing it:

c = definite_integral(sqrt(x) - x + 2, x, 2, 4)
c

-4/3*sqrt(2) + 10/3

Summing all of them will give us the result:

a + b + c

9/2

Suppose \(f\) is continous on \([a,b]\) and \(f(x) \geq 0, \forall x \in [a,b]\)

Let \(R\) denote the region under the graph of \(f\)

Let \(A\) be the area of \(R\).

We divide the interval \([a,b]\) into \(n\) pieces of width \(\dfrac{b-a}{n}\) with division points \(x_i = a + i\Delta x\)