Maths Application of Integrals

0 ≤ y ≤ x² + 1, 0 ≤ y ≤ x + 1, 1/2 ≤ x ≤ 2
Required area
= 19/24 + 5/2 = 79/24

Kindly consider the following Image

(ap-b)²+ (bp-c)²+ (cp-d)²=0 ⇒ p=b/a=c/b=d/c. a, b, c, d are in G.P.

Locus is auxiliary circle x²+y²=a²=4. (-1, √3) satisfies.

{ (x, y) ∈ R*R, x ≥ 0, 2x² ≤ y ≤ 4 − 2x}.
Required area = ∫? ¹ (4 - 2x - 2x²)dx
= [4x - x² - (2/3)x³]? ¹ = 4-1-2/3 = 7/3

y = |x| (x-1)
= { 0,   0 ≤ x < 1
{ x-1, 1 ≤ x < 2
Area = ∫? ² 2√x dx - ∫? ² (1) (1)
= [ (4x³/²)/3]? ² - 1/2 = (8√2)/3 - 1/2

$x=1$

$4x^2=8x+12$

Area $x=-\mathrm{1,3}$

$={\int }_{-1}^3?\left. [(8x+12)-4x^2\right]dx$

$={\left. [\frac{8x^2}{2}+12x-\frac{4x^3}{3}\right]}_{-1}^3$

e? F (x) = ∫ (3t²+2t+4F' (t)dt.
e? F (x)+e? F' (x) = 3x²+2x+4F' (x).
(e? -4)F' (x) = 3x²+2x-e? F (x).
F' (4) = (48+8-e? F (4)/ (e? -4).
Also F (3)=0, F (x)= (x³+x²-36)/ (e? -4) from solution. F (4)= (64+16-36)/ (e? -4) = 44/ (e? -4).
F' (4) = (56-e? (44/ (e? -4)/ (e? -4) = (56 (e? -4)-44e? )/ (e? -4)² = (12e? -224)/ (e? -4)².
α=12, β=4. α+β=16.