Question

Write the limit as a definite integral on the interval [a, b], where ci is any point in the ith subinterval. Limit Interval lim ||Δ|| → 0 n (4ci + 11) i = 1 Δxi [−8, 6]

Answer 1

Answer:

The corresponding definite integral may be written as

$\int_a^b \mathrm{(4x + 11)}\,\mathrm{d}x$

The answer of the above definite integral is

$\int_a^b \mathrm{(4x + 11)}\,\mathrm{d}x = 98$

Step-by-step explanation:

The given limit interval is

$\lim_{||\Delta|| \to 0} \sum\limits_{i=1}^n (4c_i + 11) \Delta x_i$

$[a, b] = [-8, 6]$

The corresponding definite integral may be written as

$\int_a^b \mathrm{(4x + 11)}\,\mathrm{d}x$

$\int_{-8}^6 \mathrm{(4x + 11)}\,\mathrm{d}x$

Bonus:

The definite integral may be solved as

$\int_{-8}^6 \mathrm{(4x + 11)}\,\mathrm{d}x \\\\\frac{4x^2}{2} + 11x \left \|{b=6} \atop {a=-8}} \right. \\\\2x^2 + 11x \left \|{b=6} \atop {a=-8}} \right. \\\\ 2(6^2 -(-8)^2 ) + 11(6 - (-8) \\\\2(36 - 64 ) + 11(6 + 8) \\\\2(-28 ) + 11(14) \\\\-56 +154 \\\\98$

Therefore, the answer to the integral is

$\int_a^b \mathrm{(4x + 11)}\,\mathrm{d}x = 98$