Integral practice problems

1. 1
   Integrate term by term

   Use linearity and the power antiderivative.

   $$6\frac{x^3}{3}-4\frac{x^2}{2}+3x$$
2. 2
   Simplify and add the constant

   An indefinite integral represents a family of functions.

   $$2x^3-2x^2+3x+C$$

Independent check: Differentiating the result gives 6x² - 4x + 3.

1. 1
   Apply the power rule

   The new exponent is negative two.

   $$\frac{x^{-2}}{-2}+C$$
2. 2
   Rewrite

   Express the negative power as a reciprocal.

   $$-\frac{1}{2x^2}+C$$

Independent check: Differentiating -1/(2x²) produces x to the negative three.

1. 1
   Identify the inner derivative

   The exponent 3x has derivative 3.

   $$\frac{d}{dx}(3x)=3$$
2. 2
   Compensate

   Multiply by one third before adding C.

   $$\int e^{3x}\,dx=\frac13e^{3x}+C$$

Independent check: The chain rule differentiates one third e to the 3x back to e to the 3x.

1. 1
   Use the sine antiderivative

   The base antiderivative of sine is negative cosine.

   $$\int\sin u\,du=-\cos u+C$$
2. 2
   Adjust for 2x

   Divide by the derivative of the inner angle.

   $$-\frac12\cos(2x)+C$$

Independent check: Differentiation gives sine of 2x because the two factors of 2 cancel.

1. 1
   Substitute the denominator

   Its derivative is already present.

   $$u=x-5,\qquad du=dx$$
2. 2
   Integrate the reciprocal

   Keep the domain split at x = 5.

   $$\ln|u|+C=\ln|x-5|+C$$

Independent check: Differentiating ln|x - 5| gives 1/(x - 5) wherever x is not 5.

1. 1
   Choose u

   The inner expression has derivative 2x.

   $$u=x^2+1,\qquad du=2x\,dx$$
2. 2
   Integrate in u

   The transformed integrand is one over u.

   $$\int\frac{du}{u}=\ln|u|+C$$
3. 3
   Return to x

   Because x² + 1 is positive, the absolute value is unnecessary.

   $$\ln(x^2+1)+C$$

Independent check: The logarithmic chain rule returns 2x/(x² + 1).

1. 1
   Find an antiderivative

   The derivative of x squared is 2x.

   $$F(x)=x^2$$
2. 2
   Evaluate the endpoints

   Subtract the lower value from the upper value.

   $$F(3)-F(0)=9-0=9$$

Independent check: The graph forms a triangle with base 3 and height 6, so its area is 9.

1. 1
   Identify symmetry

   Replacing x by negative x changes the sign.

   $$f(-x)=-f(x)$$
2. 2
   Use the symmetric interval

   The negative and positive contributions cancel.

   $$\int_{-1}^{1}x^3\,dx=0$$

Independent check: The antiderivative x⁴/4 has equal endpoint values at -1 and 1.

1. 1
   Confirm the order

   For 0 less than x less than 1, x is greater than x squared.

   $$x-x^2\ge0$$
2. 2
   Integrate and evaluate

   Use the power antiderivative.

   $$\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=\frac12-\frac13=\frac16$$

Independent check: Simpson's rule with the endpoints and midpoint also gives 1/6 exactly for this quadratic integrand.

1. 1
   Introduce a finite endpoint

   An improper integral is defined through a limit.

   $$\lim_{b\to\infty}\int_1^b x^{-2}\,dx$$
2. 2
   Evaluate before taking the limit

   The antiderivative is negative one over x.

   $$\lim_{b\to\infty}\left(-\frac1b+1\right)=1$$

Independent check: With u = 1/x, the improper integral becomes ∫₀¹ 1 du = 1, independently confirming both convergence and value.