Question

Evaluate 3(x^2) -4x + 2 dx from first principles (upper bound: 4, lower bound: 2 )

Answer 1

Evaluate $3(x^2) - 4x + 2 \, dx$ from first principles (upper bound: 4, lower bound: 2)

**Solution.**

We have the integral:

$$\int_{2}^{4} (3x^2 - 4x + 2) \, dx$$

Break it into three simple integrals:

$$\int_{2}^{4} 3x^2 \, dx - \int_{2}^{4} 4x \, dx + \int_{2}^{4} 2 \, dx$$

We know this formula:

$$\int x^n \, dx = \frac{x^{n+1}}{n+1}$$

At first find the value of the first integral using this formula:

$$\int_{2}^{4} 3x^2 \, dx = 3 \int_{2}^{4} x^2 \, dx = 3 \cdot \left. \frac{x^3}{3} \right|_{2}^{4} = 4^3 - 2^3 = 64 - 8 = 56$$

Then second integral:

$$\int_{2}^{4} 4x \, dx = 4 \int_{2}^{4} x \, dx = 4 \cdot \left. \frac{x^2}{2} \right|_{2}^{4} = 2 \cdot (4^2 - 2^2) = 24$$

Third integral:

$$\int_{2}^{4} 2 \, dx = 2x \bigg|_{2}^{4} = 4$$

Finally find the value of the (1) integral:

$$\int_{2}^{4} 3x^2 \, dx - \int_{2}^{4} 4x \, dx + \int_{2}^{4} 2 \, dx = 56 - 24 + 4 = 36$$

**Answer:** 36.