The problem asks us to evaluate several definite integrals. Here, we will solve the first 4 problems, labeled a, b, c, and d. a) $\int_{0}^{1} x^2 \, dx$ b) $\int_{-1}^{3} x^3 \, dx$ c) $\int_{-1}^{1} t^5 \, dt$ d) $\int_{0}^{1} x^{2/3} \, dx$

1. Problem Description

The problem asks us to evaluate several definite integrals. Here, we will solve the first 4 problems, labeled a, b, c, and d.

a) $\int_{0}^{1} x^2 \, dx$

b) $\int_{-1}^{3} x^3 \, dx$

c) $\int_{-1}^{1} t^5 \, dt$

d) $\int_{0}^{1} x^{2/3} \, dx$

2. Solution Steps

a) $\int_{0}^{1} x^2 \, dx$

We first find the antiderivative of $x^2$. Using the power rule for integration, we have:

$\int x^2 \, dx = \frac{x^{2+1}}{2+1} + C = \frac{x^3}{3} + C$.

Now, we evaluate the definite integral:

$\int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} - 0 = \frac{1}{3}$.

b) $\int_{-1}^{3} x^3 \, dx$

We first find the antiderivative of $x^3$. Using the power rule for integration, we have:

$\int x^3 \, dx = \frac{x^{3+1}}{3+1} + C = \frac{x^4}{4} + C$.

Now, we evaluate the definite integral:

$\int_{-1}^{3} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{-1}^3 = \frac{3^4}{4} - \frac{(-1)^4}{4} = \frac{81}{4} - \frac{1}{4} = \frac{80}{4} = 20$.

c) $\int_{-1}^{1} t^5 \, dt$

We first find the antiderivative of $t^5$. Using the power rule for integration, we have:

$\int t^5 \, dt = \frac{t^{5+1}}{5+1} + C = \frac{t^6}{6} + C$.

Now, we evaluate the definite integral:

$\int_{-1}^{1} t^5 \, dt = \left[ \frac{t^6}{6} \right]_{-1}^1 = \frac{1^6}{6} - \frac{(-1)^6}{6} = \frac{1}{6} - \frac{1}{6} = 0$.

Alternatively, since $t^5$ is an odd function and the limits of integration are symmetric around 0, the integral is

0.

d) $\int_{0}^{1} x^{2/3} \, dx$

We first find the antiderivative of $x^{2/3}$. Using the power rule for integration, we have:

$\int x^{2/3} \, dx = \frac{x^{(2/3)+1}}{(2/3)+1} + C = \frac{x^{5/3}}{5/3} + C = \frac{3}{5} x^{5/3} + C$.

Now, we evaluate the definite integral:

$\int_{0}^{1} x^{2/3} \, dx = \left[ \frac{3}{5} x^{5/3} \right]_0^1 = \frac{3}{5} (1)^{5/3} - \frac{3}{5} (0)^{5/3} = \frac{3}{5} (1) - \frac{3}{5} (0) = \frac{3}{5} - 0 = \frac{3}{5}$.

3. Final Answer

a) $\frac{1}{3}$

b) $20$

c) $0$

d) $\frac{3}{5}$