Evaluate the integral. ∫02(9x2−8x+6)dx

Question

Evaluate the integral.  ∫02(9x2−8x+6)dx

grbavit · Accepted Answer

Step 1
the Integration is a process of finding the anti derivative of a function. There are basically two types namely definite integral and indefinite integral.
The Definite integral consists of upper and lower limits and which is used to evaluate the area, volume, etc.. The indefinite integral is defined with no limits. The indefinite integral is used to find the anti derivatives.
Step 2
We have a definite integral, the integrand is in terms of powers of x. Therefore we have to use power rule of integration.
After integration, we need to apply the respective limit to obtain the value of integration at that particular interval.
Consider

\int_{0}^{2} (9 x^{2} - 8 x + 6) d x

.
Use the power rule of integration

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C

.
Therefore we get,

\int_{0}^{2} (9 x^{2} - 8 x + 6) d x = 9 \int_{0}^{2} x^{2} d x - 8 \int_{0}^{2} x d x + 6 \int_{0}^{2} d x

= 9 {[\frac{x^{3}}{3}]}_{0}^{2} - 8 {[\frac{x^{2}}{2}]}_{0}^{2} + 6 {[x]}_{0}^{2}

= 9 (\frac{8}{3}) - 8 (\frac{4}{2}) + 6 (2)

=24-16+12
=20
Hence we have

\int_{0}^{2} (9 x^{2} - 8 x + 6) d x = 20

.
Hence the value of the integration is 20.

Evaluate the integral. \int_{0}^{2}(9x^{2}-8x+6)dx

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