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Popular Calculus >

integral from 3 to 7 of k^2+2k-1

Solution

\int_{3}^{7} k^{2} + 2 k - 1

Solution

4 (k^{2} + 2 k - 1)

Solution steps

\int_{3}^{7} k^{2} + 2 k - 1 d x

Integral of a constant:

\int a d x = a x

= [(k^{2} + 2 k - 1) x]_{3}^{7}

Compute the boundaries:

4 (k^{2} + 2 k - 1)

= 4 (k^{2} + 2 k - 1)

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Frequently Asked Questions (FAQ)

What is the integral from 3 to 7 of k^2+2k-1 ?
The integral from 3 to 7 of k^2+2k-1 is 4(k^2+2k-1)