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If I wanted the area under a part of a curve, I could integrate with both an indefinite integral using the initial conditions or just integrate with a definite integral.

Does the constant of integration/initial conditions account for the lower bound? Is the lower bound just the initial condition?

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  • $\begingroup$ For the title question: No relation $\endgroup$ Nov 15, 2016 at 0:39
  • $\begingroup$ $$\int_a^bf(x)dx=[F(b)+c]-[F(a)+c]=F(b)-F(a)$$ $\endgroup$ Nov 15, 2016 at 0:40
  • $\begingroup$ If I am able to use both methods, there must be some sort of relationship? $\endgroup$ Nov 15, 2016 at 0:41
  • $\begingroup$ The upper bound for the definite integral is going to be the same as the value I substitute into the indefinite integral, so shouldn't the constant of integration and lower bound have a relationship? $\endgroup$ Nov 15, 2016 at 0:42
  • $\begingroup$ If I want the area under the curve, I'm having trouble what the indefinite integral does exactly. $\endgroup$ Nov 15, 2016 at 0:44

2 Answers 2

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What the indefinite integral does is compute "a" function whose derivative is the integrand. It does/can not be related directly to the area under the curve of the integrand as it does not take any arguments regarding the region whose area is to be found.

Quoting Wolfram,
Since the derivative of a constant is zero, any constant may be added to an antiderivative and will still correspond to the same integral. Another way of stating this is that the antiderivative is a non-unique inverse of the derivative. For this reason, indefinite integrals are often written in the form

$\int f(z)dz=F(z)+C$,

where C is an arbitrary constant known as the constant of integration.

The first fundamental theorem of calculus allows definite integrals to be computed in terms of indefinite integrals. In particular, this theorem states that if $F$ is the indefinite integral for a complex function $f(z)$, then

$\int_a^bf(z)dz=F(b)-F(a)$.

Also, if you want to know more about area under curves read this.

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The indefinite integral $\int f(x) dx$ is just the antiderivative of $f(x)$, or in other words the function $y = F(x)$ which is the solution to $\frac{dy}{dx} = f(x)$. However, if we are given one possible antiderivative, then we can define a whole family of valid solutions to that equation just by adding an arbitrary constant, i.e. if $y = F(x)$ is a solution, then so is $y = F(x) + C$, hence we write $\int f(x) dx = F(x) + C$ to remind us that we need additional information to uniquely determine a particular antiderivative.

Completely unrelatedly, we use the same notation for the definite integral $\int_a^b f(x) dx$ to denote the area under the curve $y = f(x)$ on the interval $x \in (a, b)$, which at least when you start learning calculus you typically approach via Riemann sums, i.e. approximate the area with rectangles then make the rectangles infinitesimally thin and see what happens.

However, the Fundamental Theorem of Calculus (FTOC) states that if $F(x)$ is an antiderivative of $f(x)$, then $\int_a^b f(x) dx = F(b) - F(a)$. You'll note that it doesn't actually matter which antiderivative we pick, as long as we don't change halfway through - if you switch to $F(x) + C$ for some value of $C$, then the definite integral becomes $(F(b) + C) - (F(a) + C) = F(b) - F(a)$, i.e. the value stays the same.

That said, we can define some particular function $G(x)$ as being the area under the function $y = f(x)$, starting from $x = a$ and going up to some arbitrary value of $x$. In other words, $G(x) = \int_a^x f(t) dt$ (note that I'm changing the variable of integration so that we don't have $x$ as both a free variable and as the variable of integration, this doesn't actually change anything). Then by FTOC, given some antiderivative $F(x)$, $G(x) = F(x) - F(a)$. But $F(a)$ is a constant, so $G(x)$ is also an antiderivative of $f(x)$, and in particular it's the antiderivative such that $G(a) = F(a) - F(a) = 0$.

So the relationship is that for a particular definite integral $\int_a^x f(t) dt$, if we use the antiderivative $G(x) = F(x) + C$, then we are implicitly setting the value $C$ such that $C = -F(a)$ and $G(a) = 0$.

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