Integral areas?

Question

Find the number b such that the line y = b divides the region bounded by the curves y=3x^2 and y = 2 into two regions with equal area.

Enter your answer either as a expression or as a decimal rounded to four places.

Answer 1

Using symmetry, we only need to consider the first quadrant.

∫2-3x^2 dx [0, √(2/3)] = 2∫b-3x^2 dx [0, √(b/3)]
Solve for b,
b = 1.2599
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Check:
∫2-3x^2 dx [0, √(2/3)] = 1.08866
2∫b-3x^2 dx [0, √(b/3)] = 1.08863 at b = 1.2599

Answer 2

Consider the rectangle bounded by the axes and the point (sqrt(2/3), 2). The curve divides this into two unequal parts, and we are interested in the upper one. The area of that part will be (2 sqrt(2/3)) - defint (0, sqrt(2/3), 3x^2 dx). The integral is x*3[0, sqrt(2/3)] or (2/3)^3/2. Evaluate this numerically, as well as the total area, to get the area above the curve. Divide that by 2 to get the desired area A. Then you want that.value of x for which x^3[0, x] = A. This will simply be the cube root of A. The resulting value of x is then plugged in to the original formula for y to get the required answer.

Answer 3

Area enclosed by y = 3x² and y = 2 is given by:-

A = ∫ x.dy where x = √y / √3

A = (1/√3) . ∫ y^(1/2) dy between limits 0 to 2

A = (1/√3) . 2/3. [y^(3/2)] limits 0 to 2

A = (2√3 / 9).( 2.√2) = (4/9).√6

(1/2) A = (2 / 9) .√6

∫ x dy = (2/9).√6 between limits 0 to b

[2 / (3√ 3)]. b^(3/2) = (2/9).(√6)

b^(3/2) = √2

b = (√2)^(3/2) = 1.681

Answer 4

integral 0 to b (3x^2)= integral b to 2 (3x^2)

b^3=4 or b=1.5874