Integration and Accumulation of Change

$\int_{\pi/4}^{\pi/2} \tan(x) dx$

$\int_{-3}^{3} e^x dx$

$\int_{1}^{\infty} \frac{1}{x^2} dx$

$\int_{0}^{1} x^2 dx$

Try to apply L'Hopital's Rule before setting up limits for the integral split.

Treat the integral as a whole and use trapezoidal rule for numerical approximation.

Split the integral into two parts, one from -infinity to some constant c, and another from c to infinity by taking limits for both parts.

Use Gaussian's exponential functions methods for direct evaluation without splitting or taking limits.

If the limit exists but is infinite, then it converges.

Convergence cannot be determined solely by limits in this context.

If the limit exists and is finite, then the improper integral converges.

The existence of this limit implies divergence regardless of its value.

Directly compute its antiderivative over $(0, \infty)$ since sine is bounded between -1 and +1 which ensures convergence without needing limits initially.

Break up the integral into intervals over which sine is continuous (like multiples of $\pi$), replace infinity with a limit variable such as $b$, evaluate each piecewise section using limits, then sum them if possible.

Utilize comparison test by comparing it to another related function which is easier to integrate over $(0, \infty)$ without setting up piecewise sections first.

Apply Laplace transform techniques on $\sin(x)$ prior to determining convergence or divergence since they offer another way of handling functions over infinite intervals directly.

The result oscillates between positive and negative values indicating inconclusive behavior for any $n$.

It results in zero if $n=1$ representing convergence.

It simplifies to $e^{-ax }$ indicating convergence regardless of $n$.

Applying L'Hôpital's rule indefinitely yields infinity suggesting divergence for all $n$.

The integral has an infinite value.

The integral approaches zero.

The integral has a finite value.

The integral oscillates between positive and negative values.

Both Type I and Type II improper integrals.

Type II improper integral.

Neither Type I nor Type II improper integrals.

Type I improper integral.

Integrate from $-\infty$ to $1$ instead and check for convergence.

Calculate the indefinite integral and substitute $x = \infty$.

Find the derivative of $\frac{1}{x^2}$ and evaluate at $x = 1$.

Evaluate the limit as $b$ approaches infinity of $\int_{1}^{b} \frac{dx}{x^2}$.

$0 < p < 1$

$p < 0$

$p = 1$

$p > 1$

The improper integral cannot be determined from this information

the improper integral depends on the lower limits of integration

The improper integral diverges

The improper integral converges