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Integral of x

Hello everyone! This article will walk you through how to take the integral of $x$. This is one of the most basic integrals, so it’s extremely important to know.

Finding the Integral

By the Fundamental Theorem of Calculus, $\displaystyle\dfrac{d}{dx} \int f(x)\,dx = f(x)$ for any continuous function $f(x)$. In other words, differention and integration are inverse operations. In fact, indefinite integrals are sometimes called antiderivatives. Therefore, $\displaystyle\int x\,dx$ is equal to some function $g(x)$ such that $\dfrac{d}{dx} g(x) = x$. Let’s think of what $g(x)$ could be. We already know that the derivative of $x^2$ is equal to $2x$, so $g(x)$ will be something similar to $x^2$. We also know that, for any constant $k$ and any differentiable function $f(x)$, $\dfrac{d}{dx} k \cdot f(x) = k \cdot \dfrac{d}{dx} f(x)$. Therefore, if we set $k$ to $\dfrac12$, we get that $g(x) = \dfrac{x^2}{2}$, and $\dfrac{d}{dx} g(x) = \dfrac12\cdot 2x = x$. Therefore, the integral of $x$ is $\dfrac{x^2}{2}$.

Another Way of Finding the Integral

We know that the integral of a function $f(x)$ from $x=a$ to $x=b$ is equal to the area under the graph of $f(x)$ from $x = a$ and $x = b$. We can use this to find the integral of $x$. If we look at a graph of $y = x$,
we can see that the area under the graph is actually an isosceles right triangle, and, for any value of $b$, the area under the curve from $x = 0$ to $x = b$ is equal to $\dfrac{b^2}{2}$, because both the base and height of the triangle are equal to $b$. By the FTC, $\displaystyle \dfrac{d}{dx}\int_0^x f(t)\,dt = f(x)$. Therefore,
\[ \dfrac{d}{dx}\left(\frac{x^2}{2}\right) = x. \]

Conclusion

You have just seen two ways to calculate the integral of $x$, and both ways can be very useful when trying to evaluate a complex integral. Sometimes, just looking for functions that have derivatives similar to your function can help, and sometimes it helps to look at the actual graph and see if you can calculate the area under it. Good luck on your future math adventures!

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