Evaluate the integral $\int_{-1}^{3} (3 - 4x) \, dx$ using the definition of the definite integral: $\int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x,$ where $\Delta x = \frac{b - a}{n}$ and $x_i = a + i \cdot \Delta x$ .

Question :

Evaluate the integral $\int_{-1}^{3} (3 - 4x) \, dx$ using the definition of the definite integral: $\int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \delta x,$ where $\delta x = \frac{b - a}{n}$ and $x_i = a + i \cdot \delta x$ .

$Evaluate the integral \int_{-1}^{3} (3 - 4x) \, dx using the definition of the | Doubtlet.com$

Solution:

Neetesh Kumar | November 17, 2024

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Step-by-step solution:

Step 1: Define the terms

For the integral $\int_{-1}^3 (3 - 4x) \, dx$ :

$a = -1$ , $b = 3$ .
$\Delta x = \dfrac{b - a}{n} = \dfrac{3 - (-1)}{n} = \dfrac{4}{n}$ .
$x_i = a + i \cdot \Delta x = -1 + i \cdot \dfrac{4}{n} = -1 + \dfrac{4i}{n}$ .

Step 2: Write the Riemann sum

The Riemann sum is:

$\displaystyle\sum_{i=1}^{n} f(x_i) \Delta x,$

where $f(x) = 3 - 4x$ . Substituting $x_i = -1 + \dfrac{4i}{n}$ into $f(x)$ :

$f(x_i) = 3 - 4\left(-1 + \dfrac{4i}{n}\right) = 3 - 4(-1) - \dfrac{16i}{n} = 3 + 4 - \dfrac{16i}{n} = 7 - \dfrac{16i}{n}.$

Thus:

$f(x_i) \Delta x = \left(7 - \dfrac{16i}{n}\right) \cdot \dfrac{4}{n}.$

Expand:

$f(x_i) \Delta x = \dfrac{28}{n} - \dfrac{64i}{n^2}.$

The Riemann sum becomes:

$\displaystyle\sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} \left(\dfrac{28}{n} - \dfrac{64i}{n^2}\right).$

Step 3: Separate the sum

Split the summation into two parts:

$\displaystyle\sum_{i=1}^{n} \left(\dfrac{28}{n} - \dfrac{64i}{n^2}\right) = \sum_{i=1}^{n} \dfrac{28}{n} - \sum_{i=1}^{n} \dfrac{64i}{n^2}.$

First term: $\displaystyle\sum_{i=1}^{n} \dfrac{28}{n} = \dfrac{28}{n} \cdot n = 28$
Second term: $\displaystyle\sum_{i=1}^{n} \dfrac{64i}{n^2}$ : Use the formula for the sum of the first $n$ integers: $\displaystyle\sum_{i=1}^{n} i = \dfrac{n(n+1)}{2}.$

Substituting: $\displaystyle\sum_{i=1}^{n} \dfrac{64i}{n^2} = \dfrac{64}{n^2} \cdot \dfrac{n(n+1)}{2} = \dfrac{64}{n^2} \cdot \dfrac{n^2 + n}{2} = \dfrac{64(n + 1)}{2n} = \dfrac{32(n + 1)}{n}.$

Step 4: Combine the terms

The Riemann sum becomes:

$\displaystyle\sum_{i=1}^{n} f(x_i) \Delta x = 28 - \frac{32(n + 1)}{n}.$

Simplify:

$\displaystyle\sum_{i=1}^{n} f(x_i) \Delta x = 28 - 32 - \frac{32}{n} = -4 - \frac{32}{n}.$

Step 5: Take the limit as $n \to \infty$

As $n \to \infty$ , the term $\frac{32}{n} \to 0$ . Thus:

$\lim_{n \to \infty} \displaystyle\sum_{i=1}^{n} f(x_i) \Delta x = -4$

Final Answer:

$\int_{-1}^{3} (3 - 4x) \, dx = -4$

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