Sum Rule for Definite Integrals

If two functions $ f(x) $ and $ g(x) $ are integrable over the interval $[a,b]$, then the definite integral of their sum equals the sum of their individual integrals: $$ \int_a^b [f(x)+g(x)] \,dx = \int_a^b f(x) \, dx + \int_a^b g(x) \, dx $$

This implies that if both $ f(x) $ and $ g(x) $ are Riemann integrable on $[a,b]$, then so is their sum, and the integral of the sum equals the sum of the integrals.

Proof
A Practical Example

Proof

Let $ f(x) $ and $ g(x) $ be Riemann integrable on $[a,b]$. Then, for every $ \epsilon > 0 $, there exist partitions $ P $ and $ Q $ such that:

$$ S(P,f) - s(P,f) < \epsilon/2 $$

$$ S(Q,g) - s(Q,g) < \epsilon/2 $$

Their union $ R = P \cup Q $ is also a valid partition, and it satisfies:

$$ S(R,f) - s(R,f) < \epsilon/2 $$

$$ S(R,g) - s(R,g) < \epsilon/2 $$

Now consider the inequality involving the upper and lower Riemann sums:

$$ s(R,f) + s(R,g) \le s(R,f+g) \le S(R,f+g) \le S(R,f) + S(R,g) $$

By the Riemann definition, the integral of $ f(x) + g(x) $ lies between these bounds:

$$ s(R,f) + s(R,g) \le \int_a^b [f(x) + g(x)] \, dx \le S(R,f) + S(R,g) $$

But also:

$$ s(R,f) + s(R,g) \le \int_a^b f(x) \, dx + \int_a^b g(x) \, dx \le S(R,f) + S(R,g) $$

Subtracting the two gives:

$$ s(R,f)+s(R,g) - [S(R,f)+S(R,g)] \le \int_a^b [f(x)+g(x)] \, dx - \left[ \int_a^b f(x) \, dx + \int_a^b g(x) \, dx \right] $$

$$ \le S(R,f)+S(R,g) - [s(R,f)+s(R,g)] $$

Note. When multiplying both sides of an inequality by a negative number (e.g., –1), the direction of the inequality reverses. So: $$ s(R,f) + s(R,g) \le \int_a^b f(x) \, dx + \int_a^b g(x) \, dx \le S(R,f) + S(R,g) $$ becomes: $$ -[s(R,f)+s(R,g)] \ge -\left[ \int_a^b f(x) \, dx + \int_a^b g(x) \, dx \right] \ge -[S(R,f)+S(R,g)] $$

Using the assumptions $ S(R,f) - s(R,f) < \epsilon/2 $ and $ S(R,g) - s(R,g) < \epsilon/2 $, we obtain:

$$ -\epsilon \le \int_a^b [f(x)+g(x)] \, dx - \left[ \int_a^b f(x) \, dx + \int_a^b g(x) \, dx \right] \le \epsilon $$

That is:

$$ \left| \int_a^b [f(x)+g(x)] \, dx - \left( \int_a^b f(x) \, dx + \int_a^b g(x) \, dx \right) \right| \le \epsilon $$

Since this holds for any $ \epsilon > 0 $, it follows that: $$ \int_a^b [f(x)+g(x)] \, dx = \int_a^b f(x) \, dx + \int_a^b g(x) \, dx $$

A Practical Example

Let the functions be:

$$ f(x) = 2x, \quad g(x) = x $$

Compute each integral over the interval $[1, 3]$:

$$ \int_1^3 2x \, dx = 3^2 - 1^2 = 8 $$

Note. The antiderivative of $ 2x $ is $ x^2 + c $, since $ D[x^2 + c] = 2x $.

And:

$$ \int_1^3 x \, dx = \frac{1}{2} \cdot 3^2 - \frac{1}{2} \cdot 1^2 = \frac{9}{2} - \frac{1}{2} = 4 $$

Note. The antiderivative of $ x $ is $ \frac{1}{2}x^2 + c $.

The sum of the two integrals is:

$$ \int_1^3 f(x) \, dx + \int_1^3 g(x) \, dx = 8 + 4 = 12 $$

Now integrate the sum directly:

$$ \int_1^3 [f(x) + g(x)] \, dx = \int_1^3 [2x + x] \, dx = \int_1^3 3x \, dx $$

Evaluating this integral:

$$ \int_1^3 3x \, dx = \frac{3}{2} \cdot 3^2 - \frac{3}{2} \cdot 1^2 = \frac{27}{2} - \frac{3}{2} = 12 $$

Note. The antiderivative of $ 3x $ is $ \frac{3}{2}x^2 + c $.

As expected, the result is the same:

$$ \int_1^3 f(x) + g(x) \, dx = \int_1^3 f(x) \, dx + \int_1^3 g(x) \, dx = 12 $$

And so on.