Calculating areas with antiderivatives

A general method to find the area under a graph \(y=f(x)\) between \(x=a\) and \(x=b\) is given by the following important theorem.

Theorem (Fundamental theorem of calculus)

Let \(f(x)\) be a continuous real-valued function on the interval \([a,b]\). Then

\[ \int_a^b f(x) \; dx = \Big[ F(x) \Big]_a^b = F(b) - F(a), \]

where \(F(x)\) is any antiderivative of \(f(x)\).

The notation \(\big[ F(x) \big]_a^b\) is just a shorthand to substitute \(x=a\) and \(x=b\) into \(F(x)\) and subtract; it is synonymous with \(F(b) - F(a)\).

Note that any antiderivative of \(f(x)\) will work in the above theorem. Indeed, if we have two different antiderivatives \(F(x)\) and \(G(x)\) of \(f(x)\), then they must differ by a constant, so \(G(x) = F(x) + c\) for some constant \(c\). Then we have to get the same answer whether we use \(F\) or \(G\), since

\begin{align*} G(b) - G(a) &= \big( F(b) + c \big) - \big( F(a) + c \big)\\ &= F(b) - F(a). \end{align*}

We give a proof of the theorem, assuming our previous statements:


y = f(x) of f(x) = x squared + 1, parabola, region from x = 0 to point on x axis marked as x shaded between parabola and x axis, this area labelled as A(x).
Detailed description of diagram

We can return to our original problem and solve it using the fundamental theorem.



\[ \int_0^5 (x^2 + 1) \; dx. \]


An antiderivative of \(x^2 + 1\) is \(\frac{1}{3} x^3 + x\), so we have

\begin{align*} \int_0^5 (x^2 + 1) \; dx &= \Bigg[ \frac{1}{3} x^3 + x \Bigg]_0^5\\ &= \Big( \frac{1}{3} \cdot 5^3 + 5 \Big) - \Big( \frac{1}{3} \cdot 0^3 + 0 \Big)\\ &= \frac{125}{3} + 5 = \frac{140}{3} = 46 \frac{2}{3}. \end{align*}
Exercise 10


\[ \int_0^8 (3x^2 + \sqrt[3]{x}) \; dx. \]

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