Respuesta :
Answer:
The corresponding definite integral may be written as
[tex]\int_a^b \mathrm{(4x + 11)}\,\mathrm{d}x[/tex]
The answer of the above definite integral is
[tex]\int_a^b \mathrm{(4x + 11)}\,\mathrm{d}x = 98[/tex]
Step-by-step explanation:
The given limit interval is
[tex]\lim_{||\Delta|| \to 0} \sum\limits_{i=1}^n (4c_i + 11) \Delta x_i[/tex]
[tex][a, b] = [-8, 6][/tex]
The corresponding definite integral may be written as
[tex]\int_a^b \mathrm{(4x + 11)}\,\mathrm{d}x[/tex]
[tex]\int_{-8}^6 \mathrm{(4x + 11)}\,\mathrm{d}x[/tex]
Bonus:
The definite integral may be solved as
[tex]\int_{-8}^6 \mathrm{(4x + 11)}\,\mathrm{d}x \\\\\frac{4x^2}{2} + 11x \left \|{b=6} \atop {a=-8}} \right. \\\\2x^2 + 11x \left \|{b=6} \atop {a=-8}} \right. \\\\ 2(6^2 -(-8)^2 ) + 11(6 - (-8) \\\\2(36 - 64 ) + 11(6 + 8) \\\\2(-28 ) + 11(14) \\\\-56 +154 \\\\98[/tex]
Therefore, the answer to the integral is
[tex]\int_a^b \mathrm{(4x + 11)}\,\mathrm{d}x = 98[/tex]
