Solve ∫ (from 0 to 2) of 54.38x7/25x wrt x

$$\int_{ 0 }^{ 2 } 54.38x \frac{ 7 }{ 25 } x d x$$

Evaluate

$$\frac{76132}{1875}\approx 40.603733333$$

Solution Steps

Multiply x and x to get x^{2}.

$$\int _{0}^{2}54.38x^{2}\times \frac{7}{25}\mathrm{d}x$$

Convert decimal number 54.38 to fraction \frac{5438}{100}. Reduce the fraction \frac{5438}{100} to lowest terms by extracting and canceling out 2.

$$\int _{0}^{2}\frac{2719}{50}x^{2}\times \frac{7}{25}\mathrm{d}x$$

Multiply \frac{2719}{50} times \frac{7}{25} by multiplying numerator times numerator and denominator times denominator.

$$\int _{0}^{2}\frac{2719\times 7}{50\times 25}x^{2}\mathrm{d}x$$

Do the multiplications in the fraction \frac{2719\times 7}{50\times 25}.

$$\int _{0}^{2}\frac{19033}{1250}x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{19033x^{2}}{1250}\mathrm{d}x$$

Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.

$$\frac{19033\int x^{2}\mathrm{d}x}{1250}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}.

$$\frac{19033x^{3}}{3750}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$\frac{19033}{3750}\times 2^{3}-\frac{19033}{3750}\times 0^{3}$$

Simplify.

$$\frac{76132}{1875}$$