Let f(x) be a positive function and
$I_1 = \int_{-\frac{1}{2}}^{1} 2x f(2x(1 - 2x))$ \, dx $\quad \text{and} \quad I_2 = \int_{-1}^{\frac{1}{2}} f(x(1 - x))$ \, dx. Then the value of $\frac{I_2}{I_1} is equal to:
Get the complete, step-by-step math solution for: "Let f(x) be a positive function and I_1 = ∫_(-(1)/(2))^(1) 2x f(2x(1 - 2x)) dx and I_2 = ∫_(-1)^((1)/(2)) f(x(1 - x)) dx. Then the value of (I_2)/(I_1...". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Evaluate
I1
using substitution
We begin by evaluating the integral
I1
. We will use a substitution to simplify the integrand. Let
t=2x
. Then
dt=2dx
. We also need to change the limits of integration. When
x=−21
,
t=2(−21)=−1
. When
x=1
,
t=2(1)=2
.
Step 2: Substitute into
I1
After substituting
t=2x
and
dx=2dt
, the integral
I1
becomes
∫−12tf(t(1−t))2dt
. We can pull the constant
21
outside the integral.
Step 3: Evaluate
I2
using substitution
Now, let's evaluate the integral
I2
. We will use a substitution to simplify the integrand. Let
u=1−x
. Then
du=−dx
, so
dx=−du
. We also need to change the limits of integration. When
x=−1
,
u=1−(−1)=2
. When
x=21
,
u=1−21=21
. Also,
x=1−u
.
Step 4: Substitute into
I2
After substituting
u=1−x
,
dx=−du
, and
x=1−u
, the integral
I2
becomes
∫221f((1−u)u)(−du)
. We can pull the negative sign outside the integral and then reverse the limits of integration, which changes the sign back to positive. This gives us
∫212f(u(1−u))du
.
Step 5: Apply property of definite integrals to
I2
Since the variable of integration is a dummy variable, we can replace
u
with
x
in the expression for
I2
. This gives us
I2=∫212f(x(1−x))dx
.
Step 6: Combine
I1
and
I2
We can split the integral
I1
into two parts using the property
∫abg(x)dx=∫acg(x)dx+∫cbg(x)dx
. We choose
c=21
to match the limits of
I2
.
Step 7: Evaluate the ratio
I1I2
Let
J=∫212f(x(1−x))dx
. Then
I2=J
. For the first part of
I1
, let
t=1−y
. Then
dt=−dy
. When
t=−1
,
y=2
. When
t=21
,
y=21
. So,
∫−121tf(t(1−t))dt=∫221(1−y)f((1−y)y)(−dy)=∫212(1−y)f(y(1−y))dy=∫212(1−x)f(x(1−x))dx
.
Therefore,
I1=21(∫212(1−x)f(x(1−x))dx+∫212xf(x(1−x))dx)=21∫212(1−x+x)f(x(1−x))dx=21∫212f(x(1−x))dx=21J
.
Thus,
I1I2=21JJ=2
.