What is the difference between a definite and an indefinite integral?

An indefinite integral returns a family of functions — the general antiderivative plus C. A definite integral has bounds and evaluates to a single number, the net accumulated area between them.

Why do I have to add "+C"?

Because the derivative of any constant is zero, infinitely many functions share the same derivative — x^2, x^2 + 5, and x^2 - 17 all differentiate to 2x. The +C represents that entire family.

How do I decide between substitution and integration by parts?

Scan the integrand for a function whose derivative also appears (up to a constant) — that is a substitution problem. If it is a product of two unrelated pieces, like x\cos x or x e^x, reach for integration by parts.

Can a definite integral be negative?

Yes. Area below the x-axis counts as negative, so a curve that dips under the axis can produce a negative or even zero total — the integral measures net accumulation, not geometric area.

Does Photo Math Solver handle definite integrals with bounds?

Yes — it reads the bounds straight from the screenshot, shows the antiderivative, and evaluates F(b) - F(a) explicitly, including substitution problems where the bounds need converting.

You can start solving integrals for free by adding the Chrome extension — try it on your own homework before deciding if you need anything more.

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Photo Math Solver Solved

\int 2x\,dx

1Apply the power rule: raise the power by 1 and divide by the new power: $\frac{2x^2}{2}$ .
2Simplify: $x^2$ .
3Add the constant of integration: $\int 2x\,dx = x^2 + C$ .

On this page01Definition 02Formulas 03Methods 04Examples 05Practice 06Mistakes 07FAQ

01Definition

What is an integral?

An integral accumulates a quantity: the area under a curve, the total distance from a velocity graph, the total cost from a marginal-cost function. Where a derivative asks "how fast is this changing right now?", an integral asks "how much has piled up in total?"

The two operations are inverses — that is the Fundamental Theorem of Calculus. An indefinite integral $\int f(x)\,dx$ recovers the family of functions whose derivative is $f(x)$ (hence the $+C$ ), while a definite integral $\int_a^b f(x)\,dx$ plugs the bounds into that antiderivative to produce one number.

Areas below the x-axis count as negative, which is why a definite integral can be zero even when the curve is nowhere flat — the accumulated area above and below cancel out.

Where you'll actually use this

Physics: integrating velocity gives distance traveled; integrating force over distance gives work done.
Engineering: areas, volumes, and centers of mass of irregular shapes all come from integrals.
Economics: total revenue accumulates from marginal revenue; consumer surplus is an area under a demand curve.
Probability: the chance a value falls in a range is the integral of its probability density.

Antiderivative: A function $F$ whose derivative is the integrand: $F′(x) = f(x)$ .
Bounds (limits of integration): The values $a$ and $b$ in $\int_a^b$ marking where accumulation starts and stops.
Integrand: The function being integrated — the expression between the $\int$ sign and $dx$ .
Net area: Area above the x-axis minus area below it; what a definite integral actually computes.

02Formulas

The formulas you need

Keep these on hand — every method below builds on them.

Power rule (integration)

\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)

Raise the power by one, divide by the new power — the reverse of differentiation.

Fundamental Theorem

\int_a^b f(x)\,dx = F(b) - F(a)

F is any antiderivative of f. Evaluate at the top bound minus the bottom bound.

u-substitution

\int f(g(x))\,g′(x)\,dx = \int f(u)\,du

The reverse chain rule: substitute u = g(x) when its derivative appears alongside.

Integration by parts

\int u\,dv = uv - \int v\,du

The reverse product rule — for integrands like x·eˣ or x·cos x.

03Methods

How to solve integrals

Pick the method that fits the problem in front of you.

When to use it: For sums of powers and standard functions where each term matches a known rule — the integral equivalent of power-rule differentiation.

1Split the integral across sums and pull constant factors out front.
2Integrate each term with the power rule: raise the exponent by 1, divide by the new exponent.
3Rewrite roots and reciprocals as powers first: $\sqrt{x} = x^{1/2}$ , $\frac{1}{x^2} = x^{-2}$ .
4Add a single $+C$ for an indefinite integral.
5For a definite integral, skip the $+C$ and evaluate $F(b) - F(a)$ instead.

04Worked examples

Integrals, solved step by step

From easy to hard — pick a problem to see its full solution.

\int 2x\,dx

1Apply the power rule: raise the power by 1 and divide by the new power: $\frac{2x^2}{2}$ .
2Simplify: $x^2$ .
3Add the constant of integration: $\int 2x\,dx = x^2 + C$ .

Try it on your own integrals homework

Photo Math Solver reads an integral straight from a screenshot — bounds included — and returns the full solution: the technique it chose, every intermediate step, and the final antiderivative or numeric value. No retyping ∫-signs and nested expressions into a calculator, and no guessing whether the problem needs substitution.

05Practice

Now you try

Work each one on paper first, then check your answer.

Problem 1

\int x^3\,dx

Problem 2

\int (2x + 3)\,dx

Problem 3

\int_0^2 3x^2\,dx

Problem 4

\int x\,(x^2 + 5)^3\,dx

Stuck on your integrals homework? Screenshot it

If an integral in your assignment involves a substitution or integration-by-parts setup that is tedious to retype, screenshot it with Photo Math Solver instead — it reads the expression straight from the page, bounds included, so you do not risk a transcription error before you even start solving.

06Common mistakes

Where points get lost

Each of these shows up on real graded work — and each has a simple fix.

The mistake	Why it happens	The fix
Forgetting the $+C$ on indefinite integrals.	The constant feels like bookkeeping, and definite-integral practice (where it cancels) trains you to drop it.	Write $+C$ the moment you integrate, not at the end. On graded work it is often a dedicated point.
Applying the power rule to a composite like $(x^2+1)^4$ without substitution.	It looks like $u^4$ , so the reflex is to write $\frac{(x^2+1)^5}{5}$ directly — but that ignores the chain rule in reverse.	Differentiate your answer to check. The correct route sets $u = x^2 + 1$ and needs a matching $2x\,dx$ in the integrand.
Losing constant factors during substitution.	When $du = 2x\,dx$ but the integrand only has $x\,dx$ , the required $\frac{1}{2}$ often silently disappears.	Solve for the exact quantity you have: $x\,dx = \frac{1}{2}\,du$ , and carry the fraction through the whole computation.
Evaluating a definite integral as $F(a) - F(b)$ instead of $F(b) - F(a)$ .	The bottom bound is written first in $\int_a^b$ , so it feels natural to plug it in first.	Top bound first, always: $F(b) - F(a)$ . A sign flip here inverts the entire answer.

Photo Math Solver shows every intermediate step, so slips like these are easy to catch before they cost you marks.

07FAQ