Question

4) Calcule a seguinte integral definida:

$\int\limits^2_0 {(2x-3)(4x^2+1)} \, dx$

Answer 1

Temos o seguinte:

$\int\limits^2_0 {(2x-3)(4x^2 + 1)} \, dx$

Assim:

$\int {(2x-3)(4x^2 + 1)} \, dx = \int {8x^3 + 2x - 12x^2 - 3} \, dx \\ \\ \int {8x^3 + 2x - 12x^2 - 3} \, dx = \int {8x^3} \, dx + \int {2x} \, dx - \int {12x^2} \, dx - \int {3} \, dx \\ \\ 8\int {x^3} \, dx + 2\int {x} \, dx - 12\int {x^2} \, dx - \int {3} \, dx \\ \\ 8 \cdot \frac{x^4}{4} + 2 \cdot \frac{x^2}{2} - 12 \cdot \frac{x^3}{3} - 3x \\ \\ 2x^4 + x^2 - 4x^3 - 3x$

Logo:

$\int\limits^2_0 {2x^4 + x^2 - 4x^3 - 3x} \, dx \\ \\ (2 \cdot 2^4 + 2^2 - 4 \cdot 2^3 - 3 \cdot 2) - (2 \cdot 0^4 + 0^2 - 4 \cdot 0^3 - 3 \cdot 0) \\ \\ = (32 + 4 - 4 \cdot 8 - 6) - 0 \\ \\ = 32 + 4 - 32 - 6 \\ \\ = -2$

Answer 2

Resposta:

Explicação passo-a-passo:

$\int\limits^2_0 {(2x-3)(4x^2+1)} \, dx \\\\=\int\limits^2_0 {(8x^3+2x-12x^2-3)} \, dx \\\\=\int\limits^2_0 {8x^3} \, dx+\int\limits^2_0 {2x} \, dx - \int\limits^2_0 {12x^2} \, dx-\int\limits^2_0 {3} \, dx\\\\ =(\frac{8}{4}x^4+\frac{2}{2}x^2-\frac{12}{3}x^3-3x)|\limits^2_0\\\\=(2x^4+x^2-4x^3-3x)|\limits^2_0\\\\=[2(2)^4 +2^2-4(2)^3-3(2)]-[2(0)^4 +0^2-4(0)^3-3(0)]\\\\=[2^5+4-2^5-6]-[0]\\\\=4-6\\\\=-2$