"Evaluate the following integrals using the Fundamental Theorem..." solved and explained

Lennie Carroll

Answered question

2021-11-08

Evaluate the following integrals using the Fundamental Theorem of Calculus.

\int_{1}^{2} \frac{z^{2} + 4}{z} d z

cyhuddwyr9

Skilled2021-11-09Added 90 answers

Step 1
the given integral is:

\int_{1}^{2} \frac{z^{2} + 4}{z} d z

we have to evaluate the given integral.
let the given integral be I.
therefore,

I = \int_{1}^{2} \frac{z^{2} + 4}{z} d z

Step 2

I = \int_{1}^{2} \frac{z^{2} + 4}{z} d z

= \int_{1}^{2} (\frac{z^{2}}{z} + \frac{4}{z}) d z

= \int_{1}^{2} (z + \frac{4}{z}) d z

= \int_{1}^{2} z d z + \int_{1}^{2} \frac{4}{z} d z

= {[\frac{z^{2}}{2}]}_{1}^{2} + 4 \int_{1}^{2} \frac{d z}{z} (\int x^{n} d x = \frac{x^{n + 1}}{n + 1})

= {[\frac{z^{2}}{2}]}_{1}^{2} + 4 {[\ln z]}_{1}^{2} (\int \frac{d x}{x} = \ln x)

= [\frac{2^{2}}{2} - \frac{1^{2}}{2}] + 4 [\ln 2 - \ln 1]

(if \int f (x) d x = F (x) t h e n \int_{a}^{b} f (x) d x = F (b) - F (a))

= [2 - \frac{1}{2}] + 4 [\ln \frac{2}{1}] (\log_{a} m - \log_{a} n = \log_{a} \frac{m}{n})

= \frac{3}{2} + 4

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