Evaluate the definite integral of x^2 + 3 with respect to x, from x = 0 to x = 1.

Step-by-step solution

1. 1

   Split the integral over the sum. Integration is linear, so the two terms can be handled independently:

   $\int_{0}^{1}\left(x^{2}+3\right)dx=\int_{0}^{1}x^{2}\,dx+\int_{0}^{1}3\,dx.$

2. 2

   Find each antiderivative with the power rule. Using $\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for $n\ne-1$:

   $\int x^{2}\,dx=\frac{x^{3}}{3},\qquad \int 3\,dx=3x.$

   So an antiderivative of the whole integrand is $F(x)=\frac{x^{3}}{3}+3x$. No constant of integration is needed for a definite integral — it would cancel in the subtraction.

3. 3

   Apply the fundamental theorem of calculus. Evaluate $F$ at the upper limit and subtract its value at the lower limit:

   $\left[\frac{x^{3}}{3}+3x\right]_{0}^{1}=F(1)-F(0).$

4. 4

   Compute the two values. At $x=1$:

   $\frac{1}{3}+3=\frac{1}{3}+\frac{9}{3}=\frac{10}{3}.$

   At $x=0$: $0+0=0$. Subtracting,

   $\int_{0}^{1}\left(x^{2}+3\right)dx=\frac{10}{3}-0=\frac{10}{3}\approx3.3333.$

5. 5

   Sanity-check with a geometric estimate. On $[0,1]$ the integrand runs from $3$ up to $4$, so the area must lie strictly between $3\times1=3$ and $4\times1=4$ — and $\tfrac{10}{3}=3.33$ does ✓. Splitting it up, the constant $3$ contributes a rectangle of area $3$ and the $x^{2}$ contributes $\tfrac13$, which is the well-known area under a unit parabola.