Question

$\int ^{x+y}_0 \: 2x \: dx$

Breakdown this integral into 2 or more than 2 integrals

No spams pls​

Answer 1

$\int ^{x+y}_0 \: 2x \: dx$

$\int_{0}^{x + y}\left( 2 x \right)dx$

First, calculate the corresponding indefinite integral: $\int{2 x d x}=x^{2}$.

According to the Fundamental Theorem of Calculus, $\int_a^b F(x) dx=f(b)-f(a)$, so just evaluate the integral at the endpoints, and that's the answer.

$\left(x^{2}\right)|_{\left(x=x + y\right)}=\left(x + y\right)^{2}$

$\left(x^{2}\right)|_{\left(x=0\right)} = 0$

$\small{\int_{0}^{x + y}\left( 2 x \right)dx=\left(x^{2}\right)|_{\left(x=x + y\right)}-\left(x^{2}\right)|_{\left(x=0\right)}=\left(x + y\right)^{2}}$

$\int_{0}^{x + y}\left( 2 x \right)dx=\left(x + y\right)^{2}$

Answer 2

$\mathbf\blue{{answer\::-}}$

∫0x+y2xdx

\int_{0}^{x + y}\left( 2 x \right)dx∫0x+y(2x)dx

First, calculate the corresponding indefinite integral: \int{2 x d x}=x^{2}∫2xdx=x2 .

According to the Fundamental Theorem of Calculus, \int_a^b F(x) dx=f(b)-f(a)∫abF(x)dx=f(b)−f(a) , so just evaluate the integral at the endpoints, and that's the answer.

\left(x^{2}\right)|_{\left(x=x + y\right)}=\left(x + y\right)^{2}(x2)∣(x=x+y)=(x+y)2

\left(x^{2}\right)|_{\left(x=0\right)} = 0(x2)∣(x=0)=0

\small{\int_{0}^{x + y}\left( 2 x \right)dx=\left(x^{2}\right)|_{\left(x=x + y\right)}-\left(x^{2}\right)|_{\left(x=0\right)}=\left(x + y\right)^{2}}∫0x+y(2x)dx=(x2)∣(x=x+y)−(x2)∣(x=0)=(x+y)2

\int_{0}^{x + y}\left( 2 x \right)dx=\left(x + y\right)^{2}∫0x+y(2x)dx=(x+y)2

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