How do you evaluate the definite integral #int (x^4-7/x^3+5/sqrtx)dx# from #[1,2]#?

Question

Christopher Agresta · Accepted Answer

The definite integral is #=-257/40+10sqrt2#

First we calculate the integral, using the formula

#intx^ndx=x^(n+1)/(n+1)+C# #(n!=-1)#

Then the integral #int_1^2(x^4-7/x^3+5/sqrtx)dx#

#=int_1^2(x^4-7x^(-3)+5x^(-1/2))dx#

#=(x^5/5-7x^(-2)/-2+5x^(1/2)/(1/2))#

#=(x^5/5+7/(2x^2)+10sqrtx)_1^2#

#=(32/5+7/8+10sqrt2)-(1/5+7/2+10)#

#=(31/5-21/8-10+10sqrt2)#

#-257/40+10sqrt2#

William Abdalla · Answer

undefined To evaluate the definite integral $\int_{1}^{2} \left(\frac{x^4 - 7}{x^3} + \frac{5}{\sqrt{x}}\right) dx$, follow these steps:

1. Rewrite the integrand as separate fractions:
   $$\frac{x^4 - 7}{x^3} + \frac{5}{\sqrt{x}}$$

2. Split the integral into two separate integrals:
   $$\int_{1}^{2} \frac{x^4 - 7}{x^3} dx + \int_{1}^{2} \frac{5}{\sqrt{x}} dx$$

3. Integrate each term separately:
   $$\text{For the first integral:} \quad \int_{1}^{2} \frac{x^4 - 7}{x^3} dx$$
   Use polynomial long division or simplify the integrand.

4. $$\text{For the second integral:} \quad \int_{1}^{2} \frac{5}{\sqrt{x}} dx$$
   Integrate $\frac{5}{\sqrt{x}}$ with respect to $x$.

5. Evaluate both integrals from $1$ to $2$.

6. Sum up the results from both integrals to find the value of the definite integral.