Question

Use the fundamental theorem of calculus to evaluate the following definite integral. Integral from 3 to 4 (2 x squared plus 9 )dx∫342x2+9 dx Integral from 3 to 4 (2 x squared plus 9 )dx∫342x2+9 dx

Answer 1

Answer:

Theorem of calculus

Step-by-step explanation:

Take the integral

$\int\limits^3_4 (2x^2+9)dx$

Integrate the sum term by term and factor out constants

$= 2 \int\limits^4_3 x^2 dx + 9 \int\limits^4_3 1 dx$

Apply the fundamental theorem of calculus.

The antiderivative of x^2 is x^3/3, while for a constat is x

$= \frac{2x^3}{3} + 9x dx$

Evaluate the limits

$= (2*\frac{(4)^3}{3}+9*4)-(2*\frac{(3)^3}{3}+9*3)$

$=\frac{101}{3}$

Answer 2

Answer:

$\int _3^42x^2+9dx=\frac{101}{3}$

Step-by-step explanation:

The Fundamental Theorem of Calculus says,

Suppose $f(x)$ is a continuous function on $[a,b]$ and also suppose that $F(x)$ is any anti-derivative for

                              $\int_{{\,a}}^{{\,b}}{{f\left( x \right)dx}} = \left. {F\left( x \right)} \right|_a^b = F\left( b \right) - F\left( a \right)$

Using the above definition, we can evaluate the definite integral $\int\limits^4_3 {2x^2+9} \, dx$

First,

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$\int _3^42x^2+9dx=\int _3^42x^2dx+\int _3^49dx$

Evaluate the integral

$\int _3^42x^2dx=2\cdot \int _3^4x^2dx\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1\\\\2\left[\frac{x^{2+1}}{2+1}\right]^4_3=2\left[\frac{x^3}{3}\right]^4_3\\\\\mathrm{Compute\:the\:boundaries}:\\\\2(\frac{4^3}{3}-\frac{3^3}{3}) =2\cdot \frac{37}{3}= \frac{74}{3}$

and the integral

$\int _3^49dx=\left[9x\right]^4_3=9(4)-9(3)=9$

Finally, we get that

$\int _3^42x^2+9dx=\frac{74}{3}+9=\frac{101}{3}$