Given that $\int_{0}^{3} f (x) d x = 6$ $int_0^3f(x)dx=6$ and $\int_{0}^{2} f (x) d x = 5$ $int_0^2f(x)dx=5$ . Show that $\int_{2}^{3} f (x) d x$ $int_2^3f(x)dx$ . How to do this?

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Given that $\int_{0}^{3} f (x) d x = 6$ $int_0^3f(x)dx=6$ and $\int_{0}^{2} f (x) d x = 5$ $int_0^2f(x)dx=5$ . Show that $\int_{2}^{3} f (x) d x$ $int_2^3f(x)dx$ . How to do this?

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Answer 1 · 2018-02-03T12:59:39

$int_2^3 f(x)\ dx=1$

Explanation:

If we consider an integral as the area under a curve, we can say that the area under the curve on the interval $[2,3]$ can be expressed as the area under the curve on the interval $[0,3]$ minus the area under the curve on the interval $[0,2]$ .

You can think of it as taking the area under $[0,3]$ and then removing the area $[0,2]$ , which would leave you with the area under $[2,3]$ .

This knowledge lets us put up the following equation of the integrals:
$int_2^3 f(x)\ dx=int_0^3 f(x)\ dx-int_0^2 f(x)\ dx$

Since we know the values for the integrals on the right hand side, we get:
$int_2^3 f(x)\ dx=6-5$

This means that the integral is equal to:
$int_2^3 f(x)\ dx=1$

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Given that $\int_{0}^{3} f (x) d x = 6$ $int_0^3f(x)dx=6$ and $\int_{0}^{2} f (x) d x = 5$ $int_0^2f(x)dx=5$ . Show that $\int_{2}^{3} f (x) d x$ $int_2^3f(x)dx$ . How to do this?

1 Answer

Explanation:

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Given that int_0^3f(x)dx=6∫30f(x)dx=6int_0^3f(x)dx=6 and int_0^2f(x)dx=5∫20f(x)dx=5int_0^2f(x)dx=5. Show that int_2^3f(x)dx∫32f(x)dxint_2^3f(x)dx. How to do this?

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Given that $\int_{0}^{3} f (x) d x = 6$ $int_0^3f(x)dx=6$ and $\int_{0}^{2} f (x) d x = 5$ $int_0^2f(x)dx=5$ . Show that $\int_{2}^{3} f (x) d x$ $int_2^3f(x)dx$ . How to do this?