If f is continuous and integral from 0 to 9 f(x)dx=4

The aim of this question is to find the integral of a given expression. Furthermore, the upper and lower limits of the integral are also given, i.e., we have a definite integral in this question.

This question is based on the concept of arithmetic. The integral tells us about the area under the curve. Furthermore, the definite integral is given in which we have upper and lower limits of the integral, therefore, we will get the exact value in the solution.

The integral of the given expression can be calculated as follows:

$\int_{0}^{3} x f (x^2) \, dx$

We will solve the expression using substitution as:

x = z  and therefore,  2 x dx = dz

By multiplying and dividing the given expression with 2, we have:

$\dfrac{1}{2} \int_{0}^{3} f(x^2) (2 x dx) \, dx$

Moreover, the integration limits are also updated, as given below:

$\int_{0}^{3} to \int_{0}^{( 3^2 )} = \int_{0}^{9}$

$\dfrac{1}{2} \int_{0}^{9} f (z) \, dz$

It is also kept in mind that by substitution, the question remained the same i.e.:

$\int_{b}^{a} f (z) \, dz = \int_{b}^{a} f (x) \, dx$

Therefore,

$\dfrac{1}{2} \int_{0}^{9} f (z) \, dz = \dfrac{1}{2} \times 4$

$\dfrac{1}{2} \times 4 = 2$

So,

$\int_{0}^{3} x f (x^2) \, dx = 2$

Numerical Results

From the solution given above, the following mathematical results are obtained:

$\int_{0}^{3} x f (x^2) \, dx = 2$

Example

If $f$ is a continuous integral $0$ to $3$ $x f(x^2) dx = 2$ find the integral $2$ to $3$ $x f (x^2) dx$.

Solution

We have all the given information, so the solution can be found as:

$\int_{2}^{3} x f (x^2) \, dx$

By substitution, we have:

$x = t$ and therefore, $2 x dx = dt$

By multiplying and dividing with 2, we have:

$\dfrac{ 1 }{ 2 } \int_{ 2 }^{ 3 } f ( x^2 ) ( 2 x dx ) \, dx$

By updating integration limits:

$\int_{2}^{3} to \int_{2^2}^{ (3^2) } = \int_{4}^{9}$

$\dfrac{1}{2} \int_{4}^{9} f (t) \, dt$

As we know, by substitution the question remained the same, therefore:

$\dfrac{1}{2} \int_{4}^{9} f (z) \, dz = \dfrac{1}{2} \times 12.6$

$\dfrac{1}{2} \times 12.6 = 6.3$

So,

$\int_{2}^{3} x f (x^2) \, dx = 6.3$

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