Find the definite integral ∫ x ^ 2 [radical (a ^ 2-x ^ 2)] DX, upper limit a, lower limit 0

Find the definite integral ∫ x ^ 2 [radical (a ^ 2-x ^ 2)] DX, upper limit a, lower limit 0

∫ x ^ 2 [radical (a ^ 2-x ^ 2)] DX
=a^2∫x^2dx-∫x^4dx
=1/3 *a^2*x^3-1/5 *x^5+c
The definite integral over 0, a is
1/3*a^5-1/5*a^5=1/15 *a^5

Find 1 DX / 1 + radical x under 4 above the definite integral f

A:
Let t = 1 + √ x, x = (t-1) ^ 2
x=1,t=2
x=4,t=3
simple form
=(2→3) ∫1/td(t-1)^2
=(2→3) ∫(1/t)*2(t-1)dt
=(2→3) 2*∫(1-1/t)dt
=(2→3) 2*(t-lnt)
=2*(3-ln3)-2(2-ln2)
=6-ln9-4+ln4
=2+ln(4/9)
=2-ln(9/4)

Find the definite integral of upper limit 4, lower limit 1, radical x (x-radical x) DX

simple form
=∫（4,1）（x＾3/2-x）dx
=2/5x＾5/2-1/2x＾2│（4,1）
=（2/5*32-1/2*16）-（2/5-1/2）
=64/5-8-2/5+1/2
=4.9
If there is anything you don't understand, you can ask,

Given that the absolute value of (root a-radical 3) 2 + radical B + 2 + C + radical 2 = 0, calculate the value of C? - 2Ab Radical B + 2 together The absolute values of C + radical 2 are connected together

∵ (root a-radical 3) 2 + radical B + 2 + ┃ C + Radix 2 ┃ = 0
∴√a-√3=0 b+2=0 c+√2=0
a=3 b=-2 c=-√2
/ / the value of C? - 2Ab
=(-√2)²-2*3*(-2)
=2+12
=14

Given the absolute value of 2004-A + a-2005 = a under the root sign, find the value of a-2004 2 + 20 under the root sign

2004-A absolute value + root sign a-2005 = a
a-2005≥0
Namely
a≥2005
therefore
The original formula is changed into
A-2004 + Radix a-2005 = a
Radix a-2005 = 2004
Square, get
a-2005=2004²
therefore
a-2004²=2005
therefore
A-2004 2 + 20 under radical
=Radical 2025
=45

(root 3 + root 2) ^ 2012 * (root 2-root 3) ^ 2013

(root 3 + root 2) ^ 2012 * (root 2-root 3) ^ 2013
=[(Gen 3 + Gen 2) (Gen 2-gen 3)] ^ 2012 * (Gen 2-gen 3)
=[2-3] ^ 2012 * (root 2-3)
=Radical 2-radical 3

The real number a satisfies | 2013 − a|+ If a − 2014 = a, then the value of a-20132 is______ .

From the meaning of the title, a-2014 ≥ 0,
∴a≥2014，
After removing the absolute number, a-2013 is obtained+
a−2014=a，
a−2014=2013，
Square the two sides, a-2014 = 20132,
∴a-20132=2014．
So the answer is: 2014

Given that X and y are real numbers, and y = Radix x-2014 + Radix 2014-x + 2013, find the value of X + y It can't be solved by a quadratic equation of one variable,

∵x-2014≥0 2014-x≥0
∴x=2014
∴y=2013
∴x+y=4027

It is known that X and y are real numbers, and y = Radix x-2014 + Radix 2014-x + 2013 Don't use a quadratic equation of one variable,

Solution y = Radix x-2014 + Radix 2014-x + 2013
Know x-2014 ≥ 0 and 2014-x ≥ 0
That is, X ≥ 2014 and X ≤ 2014
That is, x = 2014
Therefore, when x = 2014, y = √ (2014-2014) + √ (2014-2014) + 2013 = 2013
That is, x = 2014, y = 2013

Given that the real number AB satisfies the equation system a + B = 4, a ^ 2-B ^ 2 = 8, Radix 5, find the value of a ^ 2012b ^ 2013?

a²-b²=(a+b)(a-b)=8√5
a+b=4
So A-B = 2 √ 5
Add up
2a=4+2√5
a=2+√5
b=4-a=2-√5
So AB = 4-5 = - 1
So the original formula = (AB) ^ 2013
=(-1)^2013
=-1