# Calculus: Integrals I

## Definite Integrals

The previous two sections on Riemann sums gave methods for approximating the definite integral of a function over some region, known by its more familiar phrase, the area under a curve. If you were dying in anticipation - what could the exact definition of the definite integral be? Well, wait no further. Combining sums, limits, and rectangles like never before, the definite integral is defined as follows:

$$\displaystyle\int\limits_{a}^{b}f(x)\, dx = \lim\limits_{n \rightarrow \infty} \displaystyle\sum\limits_{i=1}^{n} f(x_i) \, \Delta x$$

The $x_i$ notation is used to denote the location of a sample point within each rectangular interval. In left Riemann sums, this sample point is the leftmost edge of the rectangle, and in right Riemann sums it is the rightmost edge. In the limit, the choice of location inside the region does not matter - any point will do. This formulation of the definite integral is more general than using any particular location of the sample points, although the proof of its correctness is beyond the scope of this section.

The notation of the lefthand side of the equation should now be clearer. The area of each rectangle is still present, although there is now specific notation to denote the infinitesimal width of each rectangle. The $\Delta x$ term, which conventionally denotes a discrete value, has been replaced by a $dx$ term, which denotes an infinitesimal value. Likewise, the sigma notation and sample points notation, which denote a discrete number of summed areas, have been replaced by the curvy integral symbol and just the regular $f(x)$ notation, which denote a continuous accumulation of area.

If taking limits of sums just to find areas sounds like a lot of algerbraic grunt work, you'd be right. In fact, few people compute integrals this way. Why? Because far more efficient techniques for computing integrals are right around the corner. However, sums and limits are both important aspects of mathematics (including later topics in calculus), so getting a few practice problems in is still profitable.

## Problems

1. Evaluate: $\displaystyle\int\limits_{1}^{3} x+1 \, dx$

In computing a definite integral from its definition, we need to take a few steps. The first is to calculate $\Delta x$, which isn't very hard. The next is to pick a location of a sample point inside each rectangle. Since any location will do, it helps to pick the easiest, which is often either the left-hand edge or right-hand edge. Next, we simplify the summation expression, which will likely require converting it to one or many closed-form expressions (see the practice problems in the Sums section). Finally, we take the limit of the simplified expression to get the final value.

Step 1: Calculate $\Delta x$:

$\Delta x = \dfrac{b-a}{N} \\ \Delta x = \dfrac{3-1}{N} \\ \Delta x = \dfrac{2}{N} \\$

Step 2: Determine the location of the sample point:

The right edges of the rectangles are at $1 + \dfrac{2}{N}, 1 + \dfrac{4}{N}, 1 + \dfrac{6}{N}$ etc. We can then choose $x_i = 1 + \dfrac{2i}{N}$ as the sample point.

Step 3: Substitute and simplify:

$\displaystyle\int\limits_{1}^{3} x+1 \, dx = \lim\limits_{N\rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} (x_i+1) \, \Delta x \\ \displaystyle\int\limits_{1}^{3} x+1 \, dx = \lim\limits_{N\rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} \left(1 + \dfrac{2i}{N}+1\right) \left(\dfrac{2}{N}\right) \\ \displaystyle\int\limits_{1}^{3} x+1 \, dx = \lim\limits_{N\rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} \left(2 + \dfrac{2i}{N}\right) \left(\dfrac{2}{N}\right) \\ \displaystyle\int\limits_{1}^{3} x+1 \, dx = \lim\limits_{N\rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} \dfrac{4}{N} + \dfrac{4i}{N^2} \\$

From here, we can split the sum up into two separate sums, then simplify using the closed-form sum formulas.

$\displaystyle\int\limits_{1}^{3} x+1 \, dx = \lim\limits_{N\rightarrow \infty}\left(\displaystyle\sum\limits_{i=1}^{N} \dfrac{4}{N} + \displaystyle\sum\limits_{i=1}^{N}\dfrac{4i}{N^2}\right) \\ \displaystyle\int\limits_{1}^{3} x+1 \, dx = \lim\limits_{N\rightarrow \infty}\left(\dfrac{4}{N}\displaystyle\sum\limits_{i=1}^{N} 1 + \dfrac{4}{N^2}\displaystyle\sum\limits_{i=1}^{N}i\right) \\ \displaystyle\int\limits_{1}^{3} x+1 \, dx = \lim\limits_{N\rightarrow \infty}\left(\dfrac{4}{N}(N) + \dfrac{4}{N^2}\left(\dfrac{N(N+1)}{2}\right)\right) \\ \displaystyle\int\limits_{1}^{3} x+1 \, dx = \lim\limits_{N\rightarrow \infty}\left(4 + \dfrac{2}{N^2}\left(N^2+N\right)\right) \\ \displaystyle\int\limits_{1}^{3} x+1 \, dx = \lim\limits_{N\rightarrow \infty}\left(4 + 2 + \dfrac{2}{N}\right) \\ \displaystyle\int\limits_{1}^{3} x+1 \, dx = 4 + 2 + \dfrac{2}{\infty} \\ \displaystyle\int\limits_{1}^{3} x+1 \, dx = 4 + 2 + 0 \\ \displaystyle\int\limits_{1}^{3} x+1 \, dx = 6 \\$

2. Evaluate: $\displaystyle\int\limits_{-1}^{2} -2x \, dx$

Step 1: Compute $\Delta x$

$\Delta x = \dfrac{b-a}{N} \\ \Delta x = \dfrac{2-(-1)}{N} \\ \Delta x = \dfrac{3}{N} \\$

Step 2: Determine the location of the sample point:

The right edges of the rectangles are at $-1 + \dfrac{3}{N}, -1 + \dfrac{6}{N}, -1 + \dfrac{9}{N},$ etc. We can choose $x_i = -1 + \dfrac{3i}{N}$ as the sample point.

Step 3: Substitute and simplify:

$\displaystyle\int\limits_{-1}^{2}-2x \, dx = \lim\limits_{N \rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} (-2x_i) \, \Delta x \\ \displaystyle\int\limits_{-1}^{2}-2x \, dx = \lim\limits_{N \rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} \left(-2\left(-1+\dfrac{3i}{N}\right)\right) \, \left(\dfrac{3}{N}\right) \\ \displaystyle\int\limits_{-1}^{2}-2x \, dx = \lim\limits_{N \rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} \left(2-\dfrac{6i}{N}\right) \, \left(\dfrac{3}{N}\right) \\ \displaystyle\int\limits_{-1}^{2}-2x \, dx = \lim\limits_{N \rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} \dfrac{6}{N}-\dfrac{18i}{N^2} \\$

From here, we can split the sum up into two separate sums, then simplify using the closed-form sum formulas.

$\displaystyle\int\limits_{-1}^{2}-2x \, dx = \lim\limits_{N \rightarrow \infty}\left(\displaystyle\sum\limits_{i=1}^{N} \dfrac{6}{N}-\displaystyle\sum\limits_{i=1}^{N}\dfrac{18i}{N^2}\right) \\ \displaystyle\int\limits_{-1}^{2}-2x \, dx = \lim\limits_{N \rightarrow \infty}\left(\dfrac{6}{N}\displaystyle\sum\limits_{i=1}^{N} 1 -\dfrac{18}{N^2}\displaystyle\sum\limits_{i=1}^{N}i\right) \\ \displaystyle\int\limits_{-1}^{2}-2x \, dx = \lim\limits_{N \rightarrow \infty}\left(\dfrac{6}{N}(N) -\dfrac{18}{N^2}\left(\dfrac{N(N+1)}{2}\right)\right) \\ \displaystyle\int\limits_{-1}^{2}-2x \, dx = \lim\limits_{N \rightarrow \infty}\left(6 -\dfrac{9}{N^2}(N^2+N)\right) \\ \displaystyle\int\limits_{-1}^{2}-2x \, dx = \lim\limits_{N \rightarrow \infty}\left(6 -9 - \dfrac{9}{N}\right) \\ \displaystyle\int\limits_{-1}^{2}-2x \, dx = 6 -9 - \dfrac{9}{\infty} \\ \displaystyle\int\limits_{-1}^{2}-2x \, dx = 6 -9 - 0 \\ \displaystyle\int\limits_{-1}^{2}-2x \, dx = -3 \\$

3. Evaluate: $\displaystyle\int\limits_{1}^{4} 2x^2 \, dx$

Step 1: Calculate $\Delta x$:

$\Delta x = \dfrac{b-a}{N} \\ \Delta x = \dfrac{4-1}{N} \\ \Delta x = \dfrac{3}{N} \\$

Step 2: Determine the location of the sample point:

The right edges of the rectangles are at $1 + \dfrac{3}{N}, 1 + \dfrac{6}{N}, 1 + \dfrac{9}{N},$ etc. We can choose $1 + \dfrac{3i}{N}$ as the sample point.

Step 3: Substitute and simplify:

$\displaystyle\int\limits_{1}^{4} x^2 \, dx = \lim\limits_{N \rightarrow \infty} \displaystyle\sum\limits_{i=1}^{N} x_i^2 \, \Delta x \\ \displaystyle\int\limits_{1}^{4} x^2 \, dx = \lim\limits_{N \rightarrow \infty} \displaystyle\sum\limits_{i=1}^{N} \left(1 + \dfrac{3i}{N}\right)^2 \left(\dfrac{3}{N}\right) \\ \displaystyle\int\limits_{1}^{4} x^2 \, dx = \lim\limits_{N \rightarrow \infty} \displaystyle\sum\limits_{i=1}^{N} \left(1 + \dfrac{6i}{N} + \dfrac{9i^2}{N^2}\right) \left(\dfrac{3}{N}\right) \\ \displaystyle\int\limits_{1}^{4} x^2 \, dx = \lim\limits_{N \rightarrow \infty} \displaystyle\sum\limits_{i=1}^{N} \dfrac{3}{N} + \dfrac{18i}{N^2} + \dfrac{27i^2}{N^3} \\$

From here, we can split the sum up into three separate sums, then simplify using the closed-form sum formulas.

$\displaystyle\int\limits_{1}^{4} x^2 \, dx = \lim\limits_{N \rightarrow \infty}\left( \displaystyle\sum\limits_{i=1}^{N} \dfrac{3}{N} + \displaystyle\sum\limits_{i=1}^{N}\dfrac{18i}{N^2} + \displaystyle\sum\limits_{i=1}^{N}\dfrac{27i^2}{N^3} \right) \\ \displaystyle\int\limits_{1}^{4} x^2 \, dx = \lim\limits_{N \rightarrow \infty}\left( \dfrac{3}{N}\displaystyle\sum\limits_{i=1}^{N} 1 + \dfrac{18}{N^2}\displaystyle\sum\limits_{i=1}^{N}i + \dfrac{27}{N^3}\displaystyle\sum\limits_{i=1}^{N}i^2 \right) \\ \displaystyle\int\limits_{1}^{4} x^2 \, dx = \lim\limits_{N \rightarrow \infty}\left( \dfrac{3}{N}(N) + \dfrac{18}{N^2}\left(\dfrac{N(N+1)}{2}\right) + \dfrac{27}{N^3}\left(\dfrac{N(N+1)(2N+1)}{6}\right) \right) \\ \displaystyle\int\limits_{1}^{4} x^2 \, dx = \lim\limits_{N \rightarrow \infty}\left( 3 + \dfrac{9}{N^2}\left(N^2+N)\right) + \dfrac{9}{2N^3}\left(2N^3+3N^2+N\right) \right) \\ \displaystyle\int\limits_{1}^{4} x^2 \, dx = \lim\limits_{N \rightarrow \infty}\left( 3 + 9 + \dfrac{9}{N} + 9+\dfrac{27}{2N} + \dfrac{9}{2N^2} \right) \\ \displaystyle\int\limits_{1}^{4} x^2 \, dx = 3 + 9 + \dfrac{9}{\infty} + 9 + \dfrac{27}{2\infty} + \dfrac{9}{2\infty^2} \\ \displaystyle\int\limits_{1}^{4} x^2 \, dx = 3 + 9 + 0 + 9 + 0 + 0 \\ \displaystyle\int\limits_{1}^{4} x^2 \, dx = 21 \\$

4. Evaluate: $\displaystyle\int\limits_{0}^{2} x^2 + 1 \, dx$

Step 1: Calculate $\Delta x$:

$\Delta x = \dfrac{b-a}{N} \\ \Delta x = \dfrac{2-0}{N} \\ \Delta x = \dfrac{2}{N} \\$

Step 2: Determine the location of the sample points:

The right edges of the rectangle are at $\dfrac{2}{N}, \dfrac{4}{N}, \dfrac{6}{N}$, etc. We can choose $\dfrac{2i}{N}$ as the sample point.

Step 3: Substitute and simplify:

$\displaystyle\int\limits_{0}^{2} x^2 + 1 \, dx = \lim\limits_{N \rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} \left( x_i^2 + 1 \right) \, \Delta x \\ \displaystyle\int\limits_{0}^{2} x^2 + 1 \, dx = \lim\limits_{N \rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} \left( \left(\dfrac{2i}{N}\right)^2 + 1 \right) \left(\dfrac{2}{N}\right) \\ \displaystyle\int\limits_{0}^{2} x^2 + 1 \, dx = \lim\limits_{N \rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} \left( \dfrac{4i^2}{N^2} + 1 \right) \left(\dfrac{2}{N}\right) \\ \displaystyle\int\limits_{0}^{2} x^2 + 1 \, dx = \lim\limits_{N \rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} \dfrac{8i^2}{N^3} + \dfrac{2}{N} \\$

From here, we can split the sum up into three separate sums, then simplify using the closed-form sum formulas.

$\displaystyle\int\limits_{0}^{2} x^2 + 1 \, dx = \lim\limits_{N \rightarrow \infty}\left(\displaystyle\sum\limits_{i=1}^{N} \dfrac{8i^2}{N^3} + \displaystyle\sum\limits_{i=1}^{N}\dfrac{2}{N} \right) \\ \displaystyle\int\limits_{0}^{2} x^2 + 1 \, dx = \lim\limits_{N \rightarrow \infty}\left(\dfrac{8}{N^3}\displaystyle\sum\limits_{i=1}^{N} i^2 + \dfrac{2}{N} \displaystyle\sum\limits_{i=1}^{N} 1\right) \\ \displaystyle\int\limits_{0}^{2} x^2 + 1 \, dx = \lim\limits_{N \rightarrow \infty}\left(\dfrac{8}{N^3}\left(\dfrac{N(N+1)(2N+1)}{6}\right) + \left(\dfrac{2}{N}\right) (N)\right) \\ \displaystyle\int\limits_{0}^{2} x^2 + 1 \, dx = \lim\limits_{N \rightarrow \infty}\left(\dfrac{4}{N^3}\left(\dfrac{2N^3+3N^2+N}{3}\right) + \left(\dfrac{2}{N}\right) (N)\right) \\ \displaystyle\int\limits_{0}^{2} x^2 + 1 \, dx = \lim\limits_{N \rightarrow \infty}\left(\dfrac{8}{3} + \dfrac{4}{N} + \dfrac{4}{3N^2} + 2\right) \\ \displaystyle\int\limits_{0}^{2} x^2 + 1 \, dx = \dfrac{8}{3} + \dfrac{4}{\infty} + \dfrac{4}{3\infty^2} + 2 \\ \displaystyle\int\limits_{0}^{2} x^2 + 1 \, dx = \dfrac{8}{3} + 0 + 0 + 2 \\ \displaystyle\int\limits_{0}^{2} x^2 + 1 \, dx = \dfrac{14}{3} \\$

5. Evaluate: $\displaystyle\int\limits_{-2}^{2} (x-2)(x-3) \, dx$

Step 1: Calculate $\Delta x$:

$\Delta x = \dfrac{b-a}{N} \\ \Delta x = \dfrac{2-(-2)}{N} \\ \Delta x = \dfrac{4}{N} \\$

Step 2: Determine the location of the sample point:

The right edges of the rectangles are at $-2 + \dfrac{4}{N}, -2 + \dfrac{8}{N}, -2 + \dfrac{12}{N},$ etc. We can choose $-2 + \dfrac{4i}{N}$ as the location of the sample point.

Step 3: Substitute and simplify:

$\displaystyle\int\limits_{-2}^{2} (x-2)(x-3) \, dx = \displaystyle\int\limits_{-2}^{2} x^2 - 5x + 6 \, dx \\ \displaystyle\int\limits_{-2}^{2} (x-2)(x-3) \, dx = \lim\limits_{N \rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} \left(x_i^2 - 5x_i + 6 \right)\, \Delta x \\ \displaystyle\int\limits_{-2}^{2} (x-2)(x-3) \, dx = \lim\limits_{N \rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} \left(\left(-2 + \dfrac{4i}{N}\right)^2 - 5\left(-2 + \dfrac{4i}{N}\right) + 6\right) \left(\dfrac{4}{N}\right) \\ \displaystyle\int\limits_{-2}^{2} (x-2)(x-3) \, dx = \lim\limits_{N \rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} \left(4 - \dfrac{16i}{N} + \dfrac{16i^2}{N^2} + 10 - \dfrac{20i}{N} + 6 \right)\left(\dfrac{4}{N}\right) \\ \displaystyle\int\limits_{-2}^{2} (x-2)(x-3) \, dx = \lim\limits_{N \rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} \left(20 - \dfrac{36i}{N} + \dfrac{16i^2}{N^2} \right)\left(\dfrac{4}{N}\right) \\ \displaystyle\int\limits_{-2}^{2} (x-2)(x-3) \, dx = \lim\limits_{N \rightarrow \infty}\displaystyle\sum\limits_{i=1}^{N} \dfrac{80}{N} - \dfrac{144i}{N^2} + \dfrac{64i^2}{N^3} \\$

From here, we can split the sum up into three separate sums, then simplify using the closed-form sum formulas.

$\displaystyle\int\limits_{-2}^{2} (x-2)(x-3) \, dx = \lim\limits_{N \rightarrow \infty}\left(\displaystyle\sum\limits_{i=1}^{N} \dfrac{80}{N} - \displaystyle\sum\limits_{i=1}^{N}\dfrac{144i}{N^2} + \displaystyle\sum\limits_{i=1}^{N}\dfrac{64i^2}{N^3}\right) \\ \displaystyle\int\limits_{-2}^{2} (x-2)(x-3) \, dx = \lim\limits_{N \rightarrow \infty}\left(\dfrac{80}{N}\displaystyle\sum\limits_{i=1}^{N} 1 - \dfrac{144}{N^2}\displaystyle\sum\limits_{i=1}^{N}i + \dfrac{64}{N^3}\displaystyle\sum\limits_{i=1}^{N}i^2\right) \\ \displaystyle\int\limits_{-2}^{2} (x-2)(x-3) \, dx = \lim\limits_{N \rightarrow \infty}\left(\left(\dfrac{80}{N}\right)(N) - \left(\dfrac{144}{N^2}\right)\left(\dfrac{N(N+1)}{2}\right) + \left(\dfrac{64}{N^3}\right)\left(\dfrac{N(N+1)(2N+1)}{6}\right)\right) \\ \displaystyle\int\limits_{-2}^{2} (x-2)(x-3) \, dx = \lim\limits_{N \rightarrow \infty}\left(80 - \left(\dfrac{72}{N^2}\right)\left(N^2+N\right) + \left(\dfrac{32}{N^3}\right)\left(\dfrac{2N^3+3N^2+N}{3}\right)\right) \\ \displaystyle\int\limits_{-2}^{2} (x-2)(x-3) \, dx = \lim\limits_{N \rightarrow \infty}\left(80 - 72 - \dfrac{72}{N} + \dfrac{64}{3} + \dfrac{32}{N} + \dfrac{32}{3N^2} \right) \\ \displaystyle\int\limits_{-2}^{2} (x-2)(x-3) \, dx = 80 - 72 - \dfrac{72}{\infty} + \dfrac{64}{3} + \dfrac{32}{\infty} + \dfrac{32}{3\infty^2} \\ \displaystyle\int\limits_{-2}^{2} (x-2)(x-3) \, dx = 80 - 72 - 0 + \dfrac{64}{3} + 0 + 0 \\ \displaystyle\int\limits_{-2}^{2} (x-2)(x-3) \, dx = \dfrac{88}{3} \\$