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)

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

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

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)

\frac{d}{d x} (- 5 e^{4 x})

\int cos (\frac{6}{2} t) d t

\frac{3}{s ( s ^{2} + 2 s + 5 )}

x \to 1 lim (+ (\frac{4}{2 x - 2}))

5 (x^{3} - x)^{4}

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)