Question

Consider the function g that is continuous on the interval [−10, 10] and for which integral from 0 to 10 of g of x times d times x equals 8. What is integral from 0 to 10 of open bracket g of x plus 2 close bracket times d times x equal to?

Answer 1

The value of the integral $\int\limits^{10}_0 {[g(x) + 2]} \, dx$ given that the integral  $\int\limits^{10}_0 {g(x)} \, dx$= -10, is 10.

Finding the area of the curve's undersurface is the process of integration. To do this, cover the area with as many little rectangles as possible, then add up their areas. The total gets closer to a limit that corresponds to the area under a function's curve. Finding an antiderivative of a function is the process of integration. If a function can be integrated and its integral across the domain is finite with the given bounds, then the integration is definite.

In the question, we are given that the function g is continuous on the interval [-10,10] and for which $\int\limits^{10}_0 {g(x)} \, dx$= -10.

We are asked the value of $\int\limits^{10}_0 {[g(x) + 2]} \, dx$.

The required value can be found as follows:

$\int\limits^{10}_0 {[g(x) + 2]} \, dx\\\Rightarrow \int\limits^{10}_0 {g(x)} \, dx + \int\limits^{10}_0 {2} \, dx \\\Rightarrow -10 + [2x]_{0}^{10}\\\Rightarrow -10 + [2*10 - 2*0]\\\Rightarrow -10 + 20\\\Rightarrow 10$

Thus, the value of the integral $\int\limits^{10}_0 {[g(x) + 2]} \, dx$ given that the integral  $\int\limits^{10}_0 {g(x)} \, dx$= -10, is 10.

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Answer 2

Answer:

The value of the integral

$\int\limits^{10}_0 {[g(x) + 2]} \, dx$

is 10.

The area under a curve between two fixed limits is a definite integral.

Here given that the function g that is continuous on the interval

$[-10,10]$

and for which integral from 0 to 10

$\int\limits^{10}_0 {g(x)} \, dx = - 10$

Now, to get the value of

$\int\limits^{10}_0 {[g(x) + 2]} \, dx$

$\int\limits^{10}_0 g(x)dx + \int\limits^{10}_02dx$

already know that the value of

$\int\limits^{10}_0g(x)dx = - 10$

so, Now

$- 10 + (2x)^{10}_0 \\ −10+[(2 \times 10)−(2 \times 0)]$

$−10+20 - 0$

$10$

Therefore the value of integral

$\int\limits^{10}_0 {[g(x) + 2]} \, dx$

is 10.

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