Calculus 2 Area between curves problem?

Question

What would the area be between these two curves, out to 5 decimal places?

(16/(1+x^4)) and 8x^2

Roger the Mole · Accepted Answer

Looking at their graphs, the two functions intersect at x = -1 and x = 1, and over that interval 16/(1+x^4) has the largest value.
So the desired area can be evaluated by:
x = 1 
 ∫ (16/(1+x^4)) - 8x^2 dx
x = -1

which, after a long integration, is equivalent to:
4 √2 (π + 2 coth^-1 (√2)) - 16/3
where "coth^-1" is the inverse function of the hyperbolic cotangent.
That expression evaluates to:
22.40980226154382

(If it was my problem, I would write a computer program to use the Newton-Cotes formulas or Runge–Kutta methods to do the integration numerically and avoid the messy explicit integration alluded to above.  But I have that opinion because I am more comfortable with programming than with complicated integration.)
(Later update: since the questions were coming in slowly to YA, I went ahead and did that, and I got 22.409802261583, so even beyond the requested 5 decimal places I agree with Wolfram's results, along with two other answerers here.)

For the confirmed nerds, here is the Perl program I wrote:
$z = 1/2**18;
for ($x = $z/2; $x < 1; $x += $z) {
  $y = (16/(1 + $x**4)) - 8 * $x**2;
  $s += $y;
}
print $s * $z * 2;

alex · Answer

intersect 16/(1+x^4) = 8x^2  ---> x=-1, 1
Area = ∫ (16/(1+x^4) - 8x^2) dx  , over [-1,1] = 16∫ (2/(1+x^4) - x^2) dx  , over [0,1]

2/(1+x^4) = 2/[(x^2 + x√2 + 1)(x^2 - x√2 + 1)] = (ax+b)/(x^2 + x√2 + 1) + (cx+d)/(x^2 - x√2 + 1)
find a, b ,c , d
then integrate
.....

---> Area = 16∫ (2/(1+x^4) - x^2) dx  , over [0,1] = 22.40980

Anonymous · Answer

16/(1 + (x^4)) = 8*(x^2);

Let u = x^2;

16/(1 + (u^2)) = 8u

16 = 8u*(1 + (u^2))

2 = u*(1 + (u^2));

2 = u + (u^3);

By look at this one of the answer is u = 1

(u^3) + u - 2 = 0;

(u - 1)*((u^2) + u + 2) = 0;

The only real answer is u = 1; x^2 = 1; x = +/- 1;

int_(-1)^(1) ((16/((x^4) + 1)) - (8*(x^2))) dx; this is an even function, therefore

int_0^(1) ((32/((x^4) + 1)) - (16*(x^2))) dx;

int_0^(1) 32/((x^4) + 1) dx - (16/3);

int_0^(1) sum_(n = 0)^(infinity) ((-1)^n)*(32)*(x^(4n)) dx - (16/3);

[sum_(n = 0)^(infinity) ((-1)^n)*(32)*(x^(4n + 1))/(4n + 1) from F(x = 1) - F(x = 0)] - (16/3);

(sum_(n = 0)^(infinity) (32)*((-1)^n)/(4n + 1)) - (16/3);

http://www.wolframalpha.com/input/?i=(sum_(n+%3D+0)%5E(infinity)+(32)*((-1)%5En)%2F(4n+%2B+1))+-+(16%2F3);

Captain Matticus, LandPiratesInc · Answer

Find where they intersect

16 / (1 + x^4) = 8x^2
16 = 8 * x^2 * (1 + x^4)
2 = x^2 * (1 + x^4)

x^2 = t

2 = t * (1 + t^2)
2 = t + t^3

t = 1 is a solution, so t - 1 is a root

t^3 + t - 2 = 0
(t - 1) * (at^2 + bt + c) = 0

(t - 1) * (at^2 + bt + c) = t^3 + t - 2
at^3 + bt^2 + ct - at^2 - bt - c = t^3 + 0t^2 + t - 2

a = 1
b - a = 0
c - b = 1
-c = -2

b = a
b = 1
c - 1 = 1
c = 2

(t - 1) * (t^2 + t + 2) = 0

t^2 + t + 2 = 0
t = (-1 +/- sqrt(1 - 8)) / 2
t = (-1 +/- i * sqrt(7)) / 2

t = x^2
t = 1
1 = x^2
-1 , 1 = x

Confirm that both are continuous and differentiable along the interval.  They are.  Find which one is greater

f(0) = 8 * 0^2 = 0
g(0) = 16 / (1 + 0^4) = 16

16 / (1 + x^4) > 8x^2 along the interval

Now we integrate

16 * dx / (1 + x^4) - 8 * x^2 * dx

You can follow the steps of integrating 16 * dx / (1 + x^4) here:
https://www.symbolab.com/solver/indefinite-integral-calculator/%5Cint%20%5Cleft(%5Cfrac%7B16%7D%7B1%20%2B%20x%5E%7B4%7D%7D%5Cright)dx

Let's just apply our limits

2 * sqrt(2) * (ln|x^2 + sqrt(2) * x + 1| - ln|x^2 - sqrt(2) * x + 1| - 2 * arctan(1 - sqrt(2) * x) + 2 * arctan(1 + sqrt(2) * x))

k = arctan(1 + sqrt(2)) - arctan(1 - sqrt(2))
k = arctan(a) - arctan(b)
tan(k) = tan(arctan(a) - arctan(b))
tan(k) = (tan(arctan(a)) - tan(arctan(b)) / (1 + tan(arctan(a)) * tan(arctan(b)))
tan(k) = (a - b) / (1 + a * b)
tan(k) = (1 + sqrt(2) - 1 + sqrt(2)) / (1 + (1 - 2)))
tan(k) = 2 * sqrt(2) / 0

tan(k) is undefined, so k = pi/2 (or 3pi/2)

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Calculus 2 Area between curves problem?

4 Answers