Write the limit as a definite integral on the interval where c_iis any point in the ith subinterval.

Question

Quizplus · Accepted Answer

Since the limit is a constant value, we can write it as the product of the length of the interval and the value of the function at any point within the interval. Thus, we can choose c to be any point in the ith subinterval, and we have:

Limit = 11ec1854_fd9d_80cf_acfc_1da6307b98f3_TBX8702_11

= length of ith subinterval x 11ec1854_fd9d_80cf_acfc_1da6307b98f3_TBX8702_11 at c

= (1/16) x 11ec1854_fd9d_80cf_acfc_1da6307b98f3_TBX8702_11 at c  (since the length of each subinterval is 1/16)

We can then write the definite integral as:

Integral from 11ec1854_fd9d_80d0_acfc_17fb2894432b_TBX8702_11 to 11ec1854_fd9d_a7e4_acfc_7b8f08f620ca_TBX8702_11 of (1/16) x 11ec1854_fd9d_80cf_acfc_1da6307b98f3_TBX8702_11 at c dc

This simplifies to:

(1/16) x 11ec1854_fd9d_80cf_acfc_1da6307b98f3_TBX8702_11 x Integral from 11ec1854_fd9d_80d0_acfc_17fb2894432b_TBX8702_11 to 11ec1854_fd9d_a7e4_acfc_7b8f08f620ca_TBX8702_11 of dc

= (1/16) x 11ec1854_fd9d_80cf_acfc_1da6307b98f3_TBX8702_11 x [c] from 11ec1854_fd9d_80d0_acfc_17fb2894432b_TBX8702_11 to 11ec1854_fd9d_a7e4_acfc_7b8f08f620ca_TBX8702_11

= (1/16) x 11ec1854_fd9d_80cf_acfc_1da6307b98f3_TBX8702_11 x [11ec1854_fd9d_a7e4_acfc_7b8f08f620ca_TBX8702_11 - 11ec1854_fd9d_80d0_acfc_17fb2894432b_TBX8702_11]

= [11ec1854_fd9d_a7e4_acfc_7b8f08f620ca_TBX8702_11 - 11ec1854_fd9d_80d0_acfc_17fb2894432b_TBX8702_11] x (1/16) x 11ec1854_fd9d_80cf_acfc_1da6307b98f3_TBX8702_11

= 11ec1854_fd9d_a7e4_acfc_7b8f08f620ca_TBX8702_11 - 11ec1854_fd9d_80d0_acfc_17fb2894432b_TBX8702_11

Thus, the answer is (D).

Write the Limit as a Definite Integral on the Interval