Find the integral on which the curve of y = the integral from 0 to x of (6/1+2t+t^2) dt
Can someone provide an explanation for this?
Sorry, forgot to include the "is concave up" part of the question
y = ∫[0,x] (6/1+2t+t^2) dt = ∫[0,x] 6/(t+1)^2 dt
y' = 6/(x+1)^2
y" = -12/(x+1)^3
the graph is concave up where y" > 0
That is, where x+1 < 0, or x < -1
To begin, let's first clarify the question. We are asked to find the integral of the given function: y = ∫[0 to x] (6/(1+2t+t^2)) dt.
To solve this problem, we can use the Fundamental Theorem of Calculus, which states that if F(x) is an antiderivative of f(x), then the definite integral of f(x) from a to b is F(b) - F(a), where a and b are the limits of integration.
In this case, we need to find an antiderivative for f(x) = (6/(1+2t+t^2)).
To find the antiderivative, we can try to rewrite the function to match a known form. Notice that the denominator of f(x) can be factored as (1+t)(1+t). We can make use of the method of partial fractions to simplify the integral.
Let's set up the partial fraction decomposition:
(6/(1+2t+t^2)) = A/(1+t) + B/(1+t)^2
To solve for A and B, we need to find a common denominator on the right-hand side:
(6/(1+2t+t^2)) = (A(1+t)^2 + B(1+t))/((1+t)^2)
Now, we can equate the numerators:
6 = A(1+t)^2 + B(1+t)
Expanding the equation,\[6 = A(t^2 + 2t + 1) + B(1+t)\]
Simplifying further, we get:
6 = (A + B) + (2A + B)t + At^2
Equating the coefficients of like powers of t on both sides, we get the following system of equations:
A + B = 0 (for the constant term)
2A + B = 0 (for the coefficient of t)
A = 6 (for the coefficient of t^2)
From the first equation, we get B = -A = -6. Substituting this value in the second equation, we obtain A = 6. Therefore, the partial fraction decomposition becomes:
(6/(1+2t+t^2)) = 6/(1+t) - 6/(1+t)^2
Now, we can rewrite the original integral using the partial fraction decomposition:
y = ∫[0 to x] (6/(1+2t+t^2)) dt
= ∫[0 to x] (6/(1+t) - 6/(1+t)^2) dt
Using linearity of integration, we can split this integral into two separate integrals:
y = ∫[0 to x] (6/(1+t)) dt - ∫[0 to x] (6/(1+t)^2) dt
The first integral is straightforward to evaluate using the u-substitution method. Let u = 1+t, which implies du = dt. The limits of integration will change accordingly:
y = ∫[1 to x+1] (6/u) du
= 6 ∫[1 to x+1] (1/u) du
= 6 ln|u| |[1 to x+1]
= 6 ln|x+1| - 6 ln|1|
= 6 ln|x+1|
For the second integral, we can observe that it has the form of a power rule for antiderivatives:
∫(1/(1+t)^2) dt = -(1/(1+t)).
To evaluate this integral, we can apply the initial value theorem:
y = ∫[0 to x] (6/(1+t) - 6/(1+t)^2) dt
= 6 ln|x+1| - (6/(1+x)) + (6/(1+0)) (evaluating at the limits)
= 6 ln|x+1| - (6/(1+x)) + 6
Finally, we have found the integral of the curve y = ∫[0 to x] (6/(1+2t+t^2)) dt:
y = 6 ln|x+1| - (6/(1+x)) + 6