Question

calculate (int_{1}^{10}9x dx), given the following.
(int_{1}^{10}x dx = 49.5) (int_{1}^{9}x^{2}dx=\frac{728}{3}) (int_{9}^{10}x^{2}dx=\frac{271}{3})
(int_{1}^{10}9x dx=square)

Explanation:

Step1: Use the property of integral \(\int_{a}^{b}kf(x)dx = k\int_{a}^{b}f(x)dx\)

For the integral \(\int_{1}^{10}9x dx\), since \(k = 9\) and \(f(x)=x\), we can rewrite it as \(9\int_{1}^{10}x dx\)

Step2: Substitute the given value of \(\int_{1}^{10}x dx\)

We know that \(\int_{1}^{10}x dx=49.5\)
Then \(9\int_{1}^{10}x dx=9\times49.5\)

$$9\times49.5 = 445.5$$

Answer:

\(445.5\)