You CAN Ace Calculus

 basics of integrals substitution definite integrals Some practice problems on this page require the use of integration by parts, which is a more advanced technique usually introduced in second semester calculus. If you haven't studied integration by parts yet, no worries. You can skip those practice problems and come back later, once you have covered integration by parts.

### 17Calculus Subjects Listed Alphabetically

Single Variable Calculus

 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
 Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives
 Implicit Differentiation Improper Integrals Indeterminate Forms Infinite Limits Infinite Series Infinite Series Table Infinite Series Study Techniques Infinite Series, Choosing a Test Infinite Series Exam Preparation Infinite Series Exam A Inflection Points Initial Value Problems, Laplace Transforms Integral Test Integrals Integration by Partial Fractions Integration By Parts Integration By Substitution Intermediate Value Theorem Interval of Convergence Inverse Function Derivatives Inverse Hyperbolic Derivatives Inverse Trig Derivatives
 Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization
 p-Series Parabolas (Rectangular Conics) Parabolas (Polar Conics) Parametric Equations Parametric Curves Parametric Surfaces Pinching Theorem Polar Coordinates Plane Regions, Describing Power Rule Power Series Product Rule
 Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area
 Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work

Multi-Variable Calculus

 Acceleration Vector Arc Length (Vector Functions) Arc Length Function Arc Length Parameter Conservative Vector Fields Cross Product Curl Curvature Cylindrical Coordinates
 Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem
 Lagrange Multipliers Line Integrals Partial Derivatives Partial Integrals Path Integrals Potential Functions Principal Unit Normal Vector
 Spherical Coordinates Stokes' Theorem Surface Integrals Tangent Planes Triple Integrals - Cylindrical Triple Integrals - Rectangular Triple Integrals - Spherical
 Unit Tangent Vector Unit Vectors Vector Fields Vectors Vector Functions Vector Functions Equations

Differential Equations

 Boundary Value Problems Bernoulli Equation Cauchy-Euler Equation Chebyshev's Equation Chemical Concentration Classify Differential Equations Differential Equations Euler's Method Exact Equations Existence and Uniqueness Exponential Growth/Decay
 First Order, Linear Fluids, Mixing Fourier Series Inhomogeneous ODE's Integrating Factors, Exact Integrating Factors, Linear Laplace Transforms, Solve Initial Value Problems Linear, First Order Linear, Second Order Linear Systems
 Partial Differential Equations Polynomial Coefficients Population Dynamics Projectile Motion Reduction of Order Resonance
 Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian

### Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

17calculus > integrals > definite using substitution

On this page, we assume you have studied both integration by substitution and definite integrals. When you combine the two techniques to evaluate integrals, there are some special techniques that will help you work them correctly. Those techniques are what we cover on this page.

When you have a definite integral, you need to do additional work to handle the limits of integration. Additionally, there is a mistake that many students make that could not only cause you to get an incorrect answer but also, if your instructor is paying attention, could cost you extra points. First, let's talk about the correct way to handle the limits of integration and then show you how NOT to do it.

There are basically two techniques, both of them correct but pay attention to what your instructor requires, since they may want you to use one of them and not the other. The first and best (in our opinion) way is to use the expression for u to change the limits of integration. Let's do an example to see how this works.

Evaluate $$\displaystyle{ \int_{-2}^{1}{ x \sqrt{x+2} ~dx } }$$.

$$-2\sqrt{3}/5$$

Problem Statement: Evaluate $$\displaystyle{ \int_{-2}^{1}{ x \sqrt{x+2} ~dx } }$$.
Solution: Let $$u=x+2 \to du=1~dx$$ and $$x=u-2$$.
We need change the limits of integration. The lower limit is $$x=-2$$, so we use the equation for u to get the lower limit in terms of u, i.e. $$x=-2 \to u=-2+2=0$$. We do the same thing with the upper limit of integration. $$x=1 \to u=1+2=3$$. So our integral is now
$$\displaystyle{ \int_{0}^{3}{ (u-2)\sqrt{u} ~du } = }$$ $$\displaystyle{ \int_{0}^{3}{u^{3/2}-2u^{1/2}} = }$$ $$\displaystyle{ \left[ (2/5)u^{5/2}-(4/3)u^{3/2} \right]_{0}^{3} = }$$ $$\displaystyle{ (2/5)3^{5/2} -(4/3)3^{3/2} = }$$ $$\displaystyle{ 2(3^{3/2})(3/5-2/3) = }$$ $$\displaystyle{ 6\sqrt{3}(-1/15) = -2\sqrt{3}/5 }$$

$$-2\sqrt{3}/5$$

Notice in that example, we changed the limits of integration from what they were in terms of x to their values when the variable to integration is changed to u. The limits of integration are tied to the variable of integration. So we could write the integral as $$\displaystyle{ \int_{x=-2}^{x=1}{ x \sqrt{x+2} ~dx } }$$

Okay, so that is the first way and, in our opinion, the best way to handle definite integrals when we use integration by substitution. The other way is to drop the limits of integration, evaluate the integral, go back to x's in the integral and then substitute the original limits of integration. This next example, shows how to do that.

Evaluate the same integral as the last example, but do not change the limits of integration and use correct notation.

Let $$u=x+2 \to du=1~dx$$ and $$x=u-2$$.
$$\displaystyle{ \int{ (u-2)\sqrt{u} ~du } = }$$ $$\displaystyle{ \int{u^{3/2}-2u^{1/2}} = }$$ $$\displaystyle{ (2/5)u^{5/2}-(4/3)u^{3/2} }$$
Now we convert back to x to get
$$\displaystyle{ \left[ (2/5)(x+2)^{5/2}-(4/3)(x+2)^{3/2} \right]_{-2}^{1} = -2\sqrt{3}/5 }$$

Notice in the solution to the last example, we dropped the limits of integration when we wrote the integral in terms of u. It would be incorrect to write $$\displaystyle{ \int_{-2}^{1}{ (u-2)\sqrt{u} ~du} }$$ since the limits of integration are in terms of x but the integral is in terms of u. Now, theoretically it MIGHT be okay to write $$\displaystyle{ \int_{x=-2}^{x=1}{ (u-2)\sqrt{u} ~du } }$$ but no one does that (but, of course, check with your instructor to see what they expect). So not only is the notation $$\displaystyle{ \int_{-2}^{1}{ (u-2)\sqrt{u} ~du } }$$ incorrect, but after you have completed the integration it is extremely easy to forget to go back to x's in the integral before substituting the limits of integration.

As an instructor, I ALWAYS take off points for writing the integral this way. So be extra careful. We recommend that you always change the limits of integration like we did in the first example. Here is a video that explains very well when to change the limits of integration, why instructors are so picky about it and possible mistakes that could trip you up. We highly recommend this video.

### PatrickJMT - U-Substitution: When Do I Have to Change the Limits of Integration? [8min-14secs]

video by PatrickJMT

### Practice

$$\displaystyle{ \int_{1}^{2}{\frac{2x}{\sqrt{x^2+1}}~dx} }$$

Problem Statement

$$\displaystyle{ \int_{1}^{2}{\frac{2x}{\sqrt{x^2+1}}~dx} }$$

Solution

### 887 solution video

video by PatrickJMT

Show that $$\displaystyle{ \int_{a}^{b}{f(x)~dx} = \int_{a}^{b}{f(a+b-x)~dx} }$$

Problem Statement

Show that $$\displaystyle{ \int_{a}^{b}{f(x)~dx} = \int_{a}^{b}{f(a+b-x)~dx} }$$

Solution

Use the substitution $$u=a+b-x \to du=-dx$$
$$x=a \to u=b$$
$$x=b \to u=a$$
$$\displaystyle{ \int_{a}^{b}{f(a+b-x)~dx} = \int_{b}^{a}{f(u)(-du)} = }$$ $$\displaystyle{ \int_{a}^{b}{f(u)~du} }$$
Since we can use any variable of integration, $$\displaystyle{ \int_{a}^{b}{f(u)~du} = \int_{a}^{b}{f(x)~dx} }$$
Therefore $$\displaystyle{ \int_{a}^{b}{f(x)~dx} = \int_{a}^{b}{f(a+b-x)~dx} }$$   [qed]

$$\displaystyle{ \int_{0}^{4}{x\sqrt{x^2+9}~dx} }$$

Problem Statement

$$\displaystyle{ \int_{0}^{4}{x\sqrt{x^2+9}~dx} }$$

Solution

Be careful that you don't do what she does in this video with her notation. She leaves her limits of integration in terms of x after she does the substitution with u. If you watched the last video in the discussion above, you know that doing this would result in loss of points with many instructors since it is wrong.

### 889 solution video

video by Krista King Math