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

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`

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 \\\\(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`