Integrals - Second fundamental theorem of integral calculus

Evaluate the definite integral: $\int_{2}^{3} x^2 \, dx$

Let $f(x) = x^2$. The antiderivative is $F(x) = \frac{x^3}{3}$. Applying the Second Fundamental Theorem: $\int_{2}^{3} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{2}^{3}$ $= \frac{3^3}{3} - \frac{2^3}{3}$ $= \frac{27}{3} - \frac{8}{3}$ $= \frac{19}{3}$

We find the antiderivative of $x^2$ using the power rule, then subtract the value at the lower limit $2$ from the value at the upper limit $3$.

Evaluate $\int_{0}^{\pi/4} \sec^2(x) \, dx$

The antiderivative of $\sec^2(x)$ is $\tan(x)$. Using the theorem: $\int_{0}^{\pi/4} \sec^2(x) \, dx = [\tan(x)]_{0}^{\pi/4}$ $= \tan\left(\frac{\pi}{4}\right) - \tan(0)$ $= 1 - 0 = 1$

Since the derivative of $\tan(x)$ is $\sec^2(x)$, we evaluate the tangent function at the given boundaries.

Find the value of $\int_{1}^{2} (4x^3 - 5x^2 + 6x + 9) \, dx$

First, find the antiderivative $F(x)$: $F(x) = x^4 - \frac{5x^3}{3} + 3x^2 + 9x$ Now evaluate $F(2) - F(1)$: $F(2) = 2^4 - \frac{5(2^3)}{3} + 3(2^2) + 9(2) = 16 - \frac{40}{3} + 12 + 18 = 46 - \frac{40}{3} = \frac{138 - 40}{3} = \frac{98}{3}$ $F(1) = 1^4 - \frac{5(1^3)}{3} + 3(1^2) + 9(1) = 1 - \frac{5}{3} + 3 + 9 = 13 - \frac{5}{3} = \frac{39 - 5}{3} = \frac{34}{3}$ Final result: $\begin{array}{r} 98 \\ - 34 \\ \hline 64 \end{array}$ Result $= \frac{64}{3}$

Integrate each term of the polynomial individually, then evaluate at the bounds and perform the subtraction.