Evaluate the definite integral. ∫299vdv

Question

Evaluate the definite integral.  ∫299vdv

Clis1955 · Accepted Answer

Step 1
In this question, we have a function for which we have to find the definite integral. Function is 9v with upper limit is 9 and lower limit is 2.
Step 2
The calculations have shown on whiteboard,

= \int_{2}^{9} 9 v d v = 9 \int_{2}^{9} v d v

Integration of

x^{n} = \frac{x^{n + 1}}{n + 1}

, using this we found the integration of 9v,

\int v^{n} d v = \frac{v^{n + 1}}{n + 1}

with n=1

= {[\frac{9 v^{2}}{2}]}_{2}^{9}

putting the values

= [\frac{9 {(9)}^{2}}{2} - \frac{9 {(2)}^{2}}{2}]

= [\frac{729}{2} - 18]

=346.5
Step 3
The value of definite integral of 9v with upper limit 9 and lower limit 2 is 346.5.

Geraldine Flores · Accepted Answer

Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:

\int_{a}^{b} f (x) d x = F (x) ∣_{a}^{b} = F (b) - F (a)

Step 2: In this case, f(v)=9v. Find its integral.

\frac{9 v^{2}}{2} ∣_{2}^{9}

Step 3: Since

F (v) ∣_{a}^{b} = F (b) - F (a)

, expand the above into F(9)-F(2):

\frac{9 \times 9^{2}}{2} - \frac{9 \times 2^{2}}{2}

Step 4: Simplify.

\frac{693}{2}

Evaluate the definite integral. \int_{2}^{9}9v dv

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